Chapter 13: Probability

By definition, the conditional probability of $$A$$ given that $$B$$ has occurred is

$$P(A|B) = \frac{P(A \cap B)}{P(B)}, \qquad P(B) \neq 0$$

Here $$P(B) = \frac{9}{13} \neq 0$$, so the formula applies. Substituting the given values:

$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\;4/13\;}{\;9/13\;} = \frac{4}{13} \times \frac{13}{9} = \frac{4}{9}$$

For a family with two children, writing the elder child first and using $$B$$ for boy and $$G$$ for girl, the sample space is

$$S = \{BB,\ BG,\ GB,\ GG\}$$

All four outcomes are equally likely, so each has probability $$\frac{1}{4}$$.

Let $$E$$ be the event 'both children are boys' and $$F$$ the event 'at least one child is a boy'. Then

$$E = \{BB\}, \qquad F = \{BB,\ BG,\ GB\}$$

$$E \cap F = \{BB\}$$

Hence $$P(F) = \frac{3}{4}$$ and $$P(E \cap F) = \frac{1}{4}$$.

The required conditional probability is

$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{\;1/4\;}{\;3/4\;} = \frac{1}{3}$$

One card is drawn from cards numbered $$1$$ to $$10$$, so the sample space is $$S = \{1, 2, \ldots, 10\}$$ with $$10$$ equally likely outcomes.

Let $$A$$ be the event 'the number is even' and $$B$$ the event 'the number is more than 3'. Then

$$A = \{2, 4, 6, 8, 10\}, \qquad B = \{4, 5, 6, 7, 8, 9, 10\}$$

$$A \cap B = \{4, 6, 8, 10\}$$

Hence $$P(B) = \frac{7}{10}$$ and $$P(A \cap B) = \frac{4}{10}$$.

The probability that the number is even, given that it is more than 3, is

$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\;4/10\;}{\;7/10\;} = \frac{4}{7}$$

Let $$E$$ be the event 'the chosen student studies in Class XII' and $$F$$ the event 'the chosen student is a girl'. The total number of students is $$1000$$.

Number of girls $$= 430$$, so

$$P(F) = \frac{430}{1000}$$

Number of girls studying in Class XII $$= 10\%$$ of $$430 = \frac{10}{100} \times 430 = 43$$.

These $$43$$ students are both girls and in Class XII, so

$$P(E \cap F) = \frac{43}{1000}$$

Therefore the required probability is

$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{\;43/1000\;}{\;430/1000\;} = \frac{43}{430} = \frac{1}{10} = 0.1$$

A die is thrown three times, so the sample space has $$6 \times 6 \times 6 = 216$$ equally likely outcomes, each an ordered triple.

$$B$$ is the event '6 on the first and 5 on the second throw'. The third throw can be any of the six numbers, so

$$B = \{(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)\}$$

This has $$6$$ outcomes, giving $$P(B) = \frac{6}{216}$$.

$$A$$ is the event '4 on the third throw'. An outcome common to $$A$$ and $$B$$ must have 6 first, 5 second and 4 third:

$$A \cap B = \{(6,5,4)\}, \qquad P(A \cap B) = \frac{1}{216}$$

Therefore

$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\;1/216\;}{\;6/216\;} = \frac{1}{6}$$

A die is thrown twice, so the sample space has $$6 \times 6 = 36$$ equally likely ordered pairs.

Let $$B$$ be the event 'the sum of the numbers is 6'. The pairs whose sum is $$6$$ are

$$B = \{(1,5),(2,4),(3,3),(4,2),(5,1)\}$$

so $$P(B) = \frac{5}{36}$$.

Let $$A$$ be the event 'the number 4 has appeared at least once'. The outcomes of $$B$$ that contain a $$4$$ are

$$A \cap B = \{(2,4),(4,2)\}, \qquad P(A \cap B) = \frac{2}{36}$$

Therefore the required conditional probability is

$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\;2/36\;}{\;5/36\;} = \frac{2}{5}$$

The coin is tossed. If it shows head $$(H)$$ it is tossed again; if it shows tail $$(T)$$ a die is thrown. The sample space is

$$S = \{(H,H),\ (H,T),\ (T,1),\ (T,2),\ (T,3),\ (T,4),\ (T,5),\ (T,6)\}$$

The probabilities of the outcomes are

$$P(H,H) = P(H,T) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$

$$P(T,i) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \quad \text{for } i = 1,2,\ldots,6$$

(Check: $$\frac{1}{4} + \frac{1}{4} + 6 \times \frac{1}{12} = 1$$.)

Let $$F$$ be the event 'there is at least one tail' and $$E$$ the event 'the die shows a number greater than 4'. Then

$$F = \{(H,T),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)\}$$

$$P(F) = \frac{1}{4} + 6 \times \frac{1}{12} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$$

$$E = \{(T,5),(T,6)\}, \qquad E \cap F = \{(T,5),(T,6)\}$$

$$P(E \cap F) = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6}$$

Therefore

$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{\;1/6\;}{\;3/4\;} = \frac{1}{6} \times \frac{4}{3} = \frac{2}{9}$$

We are given $$P(E) = 0.6$$, $$P(F) = 0.3$$ and $$P(E \cap F) = 0.2$$.

Finding $$P(E|F)$$: using $$P(E|F) = \dfrac{P(E \cap F)}{P(F)}$$ (since $$P(F) \neq 0$$),

$$P(E|F) = \frac{0.2}{0.3} = \frac{2}{3}$$

Finding $$P(F|E)$$: using $$P(F|E) = \dfrac{P(E \cap F)}{P(E)}$$ (since $$P(E) \neq 0$$),

$$P(F|E) = \frac{0.2}{0.6} = \frac{1}{3}$$

We are given $$P(B) = 0.5$$ and $$P(A \cap B) = 0.32$$.

Since $$P(B) \neq 0$$, the conditional probability formula applies:

$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.32}{0.5} = \frac{32}{50} = 0.64$$

We are given $$2P(A) = P(B) = \dfrac{5}{13}$$ and $$P(A|B) = \dfrac{2}{5}$$.

Step 1 — find $$P(A)$$ and $$P(B)$$.

$$P(B) = \frac{5}{13}, \qquad 2P(A) = \frac{5}{13} \ \Rightarrow\ P(A) = \frac{5}{26}$$

Step 2 — find $$P(A \cap B)$$. From $$P(A|B) = \dfrac{P(A \cap B)}{P(B)}$$,

$$P(A \cap B) = P(A|B) \times P(B) = \frac{2}{5} \times \frac{5}{13} = \frac{2}{13}$$

Step 3 — apply the addition theorem.

$$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13}$$

Taking LCM $$26$$: $$\dfrac{5}{26} + \dfrac{10}{26} - \dfrac{4}{26} = \dfrac{11}{26}$$

A die is thrown three times, so the sample space has $$6^3 = 216$$ equally likely ordered triples.

$$F$$ = '6 on the first toss and 5 on the second toss'. The third toss is unrestricted, so

$$F = \{(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)\}$$

This has $$6$$ outcomes, so $$P(F) = \dfrac{6}{216}$$.

$$E$$ = '4 on the third toss'. An outcome in both $$E$$ and $$F$$ must be $$(6,5,4)$$:

$$E \cap F = \{(6,5,4)\}, \qquad P(E \cap F) = \frac{1}{216}$$

Therefore

$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{\;1/216\;}{\;6/216\;} = \frac{1}{6}$$

Let $$M$$, $$F$$, $$S$$ denote mother, father and son. The three line up in a row, so the sample space consists of $$3! = 6$$ equally likely arrangements (read left to right):

$$S = \{MFS,\ MSF,\ FMS,\ FSM,\ SMF,\ SFM\}$$

Event $$E$$ = 'the son stands at one of the two ends', i.e. the letter $$S$$ is first or last. Checking each arrangement:

$$E = \{MFS,\ FMS,\ SMF,\ SFM\}$$

(In $$MFS$$ and $$FMS$$ the son is at the right end; in $$SMF$$ and $$SFM$$ the son is at the left end. The arrangements $$MSF$$ and $$FSM$$ are excluded, since there the son stands in the middle.)

Event $$F$$ = 'the father is in the middle', i.e. $$F$$ is the second letter:

$$F = \{MFS,\ SFM\}$$

$$E \cap F$$ = 'son on an end and father in the middle' $$= \{MFS,\ SFM\}$$

So $$P(F) = \dfrac{2}{6}$$ and $$P(E \cap F) = \dfrac{2}{6}$$. Hence

$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{\;2/6\;}{\;2/6\;} = 1$$

The total number of questions in the bank is

$$300 + 200 + 500 + 400 = 1400$$

Let $$E$$ = 'the question is easy' and $$M$$ = 'the question is a multiple choice question (MCQ)'.

Number of MCQs $$= 500 \text{ (easy)} + 400 \text{ (difficult)} = 900$$, so

$$P(M) = \frac{900}{1400}$$

Number of questions that are both easy and MCQ $$= 500$$, so

$$P(E \cap M) = \frac{500}{1400}$$

Therefore the probability that the question is easy, given that it is an MCQ, is

$$P(E|M) = \frac{P(E \cap M)}{P(M)} = \frac{\;500/1400\;}{\;900/1400\;} = \frac{500}{900} = \frac{5}{9}$$

Two dice are thrown, so the sample space has $$6 \times 6 = 36$$ equally likely ordered pairs.

Let $$B$$ = 'the two numbers are different'. The outcomes with equal numbers are $$(1,1),(2,2),\ldots,(6,6)$$ — that is $$6$$ outcomes. So the number of outcomes with different numbers is $$36 - 6 = 30$$, giving

$$P(B) = \frac{30}{36}$$

Let $$A$$ = 'the sum of the numbers is 4'. The outcomes with sum $$4$$ are $$\{(1,3),(2,2),(3,1)\}$$. Of these, only $$(1,3)$$ and $$(3,1)$$ have different numbers, so

$$A \cap B = \{(1,3),(3,1)\}, \qquad P(A \cap B) = \frac{2}{36}$$

Therefore

$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\;2/36\;}{\;30/36\;} = \frac{2}{30} = \frac{1}{15}$$

A die is thrown first. If a multiple of $$3$$ (that is $$3$$ or $$6$$) comes up, the die is thrown again; if any other number ($$1, 2, 4, 5$$) comes up, a coin is tossed. The sample space is

$$S = \{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6),$$
$$(1,H),(1,T),(2,H),(2,T),(4,H),(4,T),(5,H),(5,T)\}$$

Let $$F$$ = 'at least one die shows a 3'. This needs the first throw to be $$3$$ (then the second throw is anything), or the first throw to be $$6$$ with the second throw a $$3$$:

$$F = \{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,3)\}$$

Let $$E$$ = 'the coin shows a tail'. A coin is tossed only when the first throw is $$1, 2, 4$$ or $$5$$:

$$E = \{(1,T),(2,T),(4,T),(5,T)\}$$

Event $$F$$ occurs only when the die is thrown twice, while event $$E$$ occurs only when a coin is tossed. These cannot happen together, so

$$E \cap F = \phi \quad \Rightarrow \quad P(E \cap F) = 0$$

Therefore

$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{0}{P(F)} = 0$$

The conditional probability $$P(A|B)$$ is defined by

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

This formula is valid only when $$P(B) \neq 0$$, since division by zero is not allowed.

Here we are given $$P(B) = 0$$. Therefore the denominator is zero and $$P(A|B)$$ is not defined.

Hence the correct option is (c).

Write the two conditional probabilities using their definitions:

$$P(A|B) = \frac{P(A \cap B)}{P(B)}, \qquad P(B|A) = \frac{P(A \cap B)}{P(A)}$$

We are given $$P(A|B) = P(B|A)$$, so

$$\frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B)}{P(A)}$$

Since $$P(A \cap B)$$ is a common (non-zero) factor, dividing both sides by $$P(A \cap B)$$ gives

$$\frac{1}{P(B)} = \frac{1}{P(A)} \quad \Rightarrow \quad P(A) = P(B)$$

This does not force $$A = B$$ or $$A \subset B$$ or $$A \cap B = \phi$$; only the probabilities are equal. Hence the correct option is (d).

The urn contains $$10 + 5 = 15$$ balls. Let

$$E$$ = 'the first ball drawn is black', $$F$$ = 'the second ball drawn is black'.

We want $$P(E \cap F)$$, the probability that both balls are black. By the multiplication theorem,

$$P(E \cap F) = P(E)\, P(F|E)$$

First draw: there are $$10$$ black balls out of $$15$$, so

$$P(E) = \frac{10}{15}$$

Second draw (given the first was black): the urn now has $$14$$ balls of which $$9$$ are black, so

$$P(F|E) = \frac{9}{14}$$

Therefore

$$P(E \cap F) = \frac{10}{15} \times \frac{9}{14} = \frac{90}{210} = \frac{3}{7}$$

A pack has $$52$$ cards, including $$4$$ kings and $$4$$ aces. Define the events

$$K_1$$ : first card is a king, $$K_2$$ : second card is a king, $$A$$ : third card is an ace.

We want $$P(K_1 \cap K_2 \cap A)$$, and by the multiplication theorem for three events,

$$P(K_1 \cap K_2 \cap A) = P(K_1)\, P(K_2|K_1)\, P(A|K_1 \cap K_2)$$

First card: $$4$$ kings out of $$52$$ ⇒ $$P(K_1) = \dfrac{4}{52}$$.

Second card (one king already removed): $$3$$ kings out of $$51$$ ⇒ $$P(K_2|K_1) = \dfrac{3}{51}$$.

Third card (two kings removed, all $$4$$ aces still present): $$4$$ aces out of $$50$$ ⇒ $$P(A|K_1 \cap K_2) = \dfrac{4}{50}$$.

Therefore

$$P(K_1 \cap K_2 \cap A) = \frac{4}{52} \times \frac{3}{51} \times \frac{4}{50} = \frac{48}{132600} = \frac{2}{5525}$$

The sample space is $$\{1,2,3,4,5,6\}$$ with $$6$$ equally likely outcomes.

$$E$$ = 'the number is a multiple of 3' $$= \{3,6\} \Rightarrow P(E) = \dfrac{2}{6} = \dfrac{1}{3}$$

$$F$$ = 'the number is even' $$= \{2,4,6\} \Rightarrow P(F) = \dfrac{3}{6} = \dfrac{1}{2}$$

$$E \cap F = \{6\} \Rightarrow P(E \cap F) = \dfrac{1}{6}$$

Now compute $$P(E)\,P(F)$$:

$$P(E)\, P(F) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$$

Since $$P(E \cap F) = \dfrac{1}{6} = P(E)\,P(F)$$, the events $$E$$ and $$F$$ are independent.

An unbiased die is thrown twice, so the sample space has $$6 \times 6 = 36$$ equally likely ordered pairs.

$$A$$ = 'odd number on the first throw'. The first number is one of $$1,3,5$$ (3 choices) and the second is any of $$6$$, so $$A$$ has $$3 \times 6 = 18$$ outcomes:

$$P(A) = \frac{18}{36} = \frac{1}{2}$$

$$B$$ = 'odd number on the second throw'. Similarly $$B$$ has $$6 \times 3 = 18$$ outcomes:

$$P(B) = \frac{18}{36} = \frac{1}{2}$$

$$A \cap B$$ = 'odd on both throws'. This has $$3 \times 3 = 9$$ outcomes:

$$P(A \cap B) = \frac{9}{36} = \frac{1}{4}$$

Now $$P(A)\,P(B) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$$.

Since $$P(A \cap B) = \dfrac{1}{4} = P(A)\,P(B)$$, the events $$A$$ and $$B$$ are independent.

Three coins are tossed, so the sample space has $$2^3 = 8$$ equally likely outcomes:

$$S = \{HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT\}$$

$$E$$ = 'three heads or three tails' $$= \{HHH,\ TTT\} \Rightarrow P(E) = \dfrac{2}{8} = \dfrac{1}{4}$$

$$F$$ = 'at least two heads' $$= \{HHH,\ HHT,\ HTH,\ THH\} \Rightarrow P(F) = \dfrac{4}{8} = \dfrac{1}{2}$$

$$G$$ = 'at most two heads' = all outcomes except $$HHH$$ $$\Rightarrow P(G) = \dfrac{7}{8}$$

Pair (E, F): $$E \cap F = \{HHH\}$$, so $$P(E \cap F) = \dfrac{1}{8}$$.
$$P(E)\,P(F) = \dfrac{1}{4} \times \dfrac{1}{2} = \dfrac{1}{8} = P(E \cap F)$$ ⇒ independent.

Pair (E, G): $$E \cap G = \{TTT\}$$, so $$P(E \cap G) = \dfrac{1}{8}$$.
$$P(E)\,P(G) = \dfrac{1}{4} \times \dfrac{7}{8} = \dfrac{7}{32} \neq \dfrac{1}{8}$$ ⇒ dependent.

Pair (F, G): $$F \cap G$$ = 'exactly two heads' $$= \{HHT,\ HTH,\ THH\}$$, so $$P(F \cap G) = \dfrac{3}{8}$$.
$$P(F)\,P(G) = \dfrac{1}{2} \times \dfrac{7}{8} = \dfrac{7}{16} \neq \dfrac{3}{8}$$ ⇒ dependent.

Since $$E$$ and $$F$$ are independent, by definition

$$P(E \cap F) = P(E)\,P(F) \qquad \text{...(1)}$$

We must show that $$E$$ and $$F'$$ are independent, i.e. that $$P(E \cap F') = P(E)\,P(F')$$.

The event $$E$$ can be split according to whether $$F$$ occurs or not. The sets $$E \cap F$$ and $$E \cap F'$$ are mutually exclusive and their union is $$E$$:

$$E = (E \cap F) \cup (E \cap F')$$

Therefore

$$P(E) = P(E \cap F) + P(E \cap F')$$

$$\Rightarrow \quad P(E \cap F') = P(E) - P(E \cap F)$$

Using (1), $$P(E \cap F) = P(E)\,P(F)$$:

$$P(E \cap F') = P(E) - P(E)\,P(F) = P(E)\big[\,1 - P(F)\,\big] = P(E)\,P(F')$$

Since $$P(E \cap F') = P(E)\,P(F')$$, the events $$E$$ and $$F'$$ are independent. Hence proved.

'At least one of $$A$$ and $$B$$ occurs' is the event $$A \cup B$$. So we must show

$$P(A \cup B) = 1 - P(A')\,P(B')$$

The complement of $$A \cup B$$ is, by De Morgan's law,

$$(A \cup B)' = A' \cap B'$$

Hence

$$P(A \cup B) = 1 - P\big((A \cup B)'\big) = 1 - P(A' \cap B')$$

Now, since $$A$$ and $$B$$ are independent, their complements $$A'$$ and $$B'$$ are also independent. (If $$A, B$$ are independent, so are $$A, B'$$, and likewise $$A', B'$$.) Therefore

$$P(A' \cap B') = P(A')\,P(B')$$

Substituting,

$$P(A \cup B) = 1 - P(A')\,P(B')$$

which is the required result. Hence proved.

For two independent events $$A$$ and $$B$$, the probability of their joint occurrence is the product of their probabilities:

$$P(A \cap B) = P(A)\,P(B)$$

Substituting $$P(A) = \dfrac{3}{5}$$ and $$P(B) = \dfrac{1}{5}$$:

$$P(A \cap B) = \frac{3}{5} \times \frac{1}{5} = \frac{3}{25}$$

A pack of $$52$$ cards contains $$26$$ black cards (clubs and spades). Let

$$A$$ = 'the first card is black', $$B$$ = 'the second card is black'.

We want $$P(A \cap B)$$, which by the multiplication theorem is

$$P(A \cap B) = P(A)\,P(B|A)$$

First card: $$26$$ black out of $$52$$ ⇒ $$P(A) = \dfrac{26}{52}$$.

Second card (one black card already removed): $$25$$ black out of $$51$$ ⇒ $$P(B|A) = \dfrac{25}{51}$$.

Therefore

$$P(A \cap B) = \frac{26}{52} \times \frac{25}{51} = \frac{1}{2} \times \frac{25}{51} = \frac{25}{102}$$

The box has $$15$$ oranges: $$12$$ good and $$3$$ bad. The box is approved only if all three oranges drawn (without replacement) are good. Let

$$A$$ = '1st orange good', $$B$$ = '2nd orange good', $$C$$ = '3rd orange good'.

The required probability is $$P(A \cap B \cap C) = P(A)\,P(B|A)\,P(C|A \cap B)$$.

First draw: $$12$$ good out of $$15$$ ⇒ $$P(A) = \dfrac{12}{15}$$.

Second draw (one good removed): $$11$$ good out of $$14$$ ⇒ $$P(B|A) = \dfrac{11}{14}$$.

Third draw (two good removed): $$10$$ good out of $$13$$ ⇒ $$P(C|A \cap B) = \dfrac{10}{13}$$.

Therefore the probability the box is approved is

$$P(A \cap B \cap C) = \frac{12}{15} \times \frac{11}{14} \times \frac{10}{13} = \frac{1320}{2730} = \frac{44}{91}$$

A coin and a die are tossed together, so the sample space has $$2 \times 6 = 12$$ equally likely outcomes:

$$S = \{(H,1),(H,2),\ldots,(H,6),(T,1),(T,2),\ldots,(T,6)\}$$

$$A$$ = 'head appears on the coin' $$= \{(H,1),(H,2),(H,3),(H,4),(H,5),(H,6)\}$$, so $$P(A) = \dfrac{6}{12} = \dfrac{1}{2}$$.

$$B$$ = '3 on the die' $$= \{(H,3),(T,3)\}$$, so $$P(B) = \dfrac{2}{12} = \dfrac{1}{6}$$.

$$A \cap B = \{(H,3)\}$$, so $$P(A \cap B) = \dfrac{1}{12}$$.

Now $$P(A)\,P(B) = \dfrac{1}{2} \times \dfrac{1}{6} = \dfrac{1}{12}$$.

Since $$P(A \cap B) = \dfrac{1}{12} = P(A)\,P(B)$$, the events $$A$$ and $$B$$ are independent.

The die shows one of $$\{1,2,3,4,5,6\}$$, all equally likely. Numbers $$1,2,3$$ are red and $$4,5,6$$ are green.

$$A$$ = 'the number is even' $$= \{2,4,6\} \Rightarrow P(A) = \dfrac{3}{6} = \dfrac{1}{2}$$

$$B$$ = 'the number is red' $$= \{1,2,3\} \Rightarrow P(B) = \dfrac{3}{6} = \dfrac{1}{2}$$

$$A \cap B$$ = 'even and red' $$= \{2\} \Rightarrow P(A \cap B) = \dfrac{1}{6}$$

Now $$P(A)\,P(B) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$$.

Since $$P(A \cap B) = \dfrac{1}{6} \neq \dfrac{1}{4} = P(A)\,P(B)$$, the events $$A$$ and $$B$$ are not independent.

Two events are independent if and only if $$P(E \cap F) = P(E)\,P(F)$$.

Compute the product:

$$P(E)\,P(F) = \frac{3}{5} \times \frac{3}{10} = \frac{9}{50}$$

The given value of the intersection is

$$P(E \cap F) = \frac{1}{5} = \frac{10}{50}$$

Since $$P(E \cap F) = \dfrac{10}{50} \neq \dfrac{9}{50} = P(E)\,P(F)$$, the events $$E$$ and $$F$$ are not independent.

'Not $$A$$ and not $$B$$' is the event $$A' \cap B'$$. By De Morgan's law,

$$A' \cap B' = (A \cup B)'$$

so

$$P(\text{not } A \text{ and not } B) = P(A' \cap B') = 1 - P(A \cup B)$$

Step 1 — find $$P(A \cup B)$$ using the addition theorem:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8}$$

Taking LCM $$8$$: $$\dfrac{2}{8} + \dfrac{4}{8} - \dfrac{1}{8} = \dfrac{5}{8}$$

Step 2 — take the complement:

$$P(A' \cap B') = 1 - P(A \cup B) = 1 - \frac{5}{8} = \frac{3}{8}$$

'Not $$A$$ or not $$B$$' is the event $$A' \cup B'$$. By De Morgan's law,

$$A' \cup B' = (A \cap B)'$$

So

$$P(\text{not } A \text{ or not } B) = P(A' \cup B') = 1 - P(A \cap B)$$

Given that this equals $$\dfrac{1}{4}$$,

$$\frac{1}{4} = 1 - P(A \cap B) \quad \Rightarrow \quad P(A \cap B) = 1 - \frac{1}{4} = \frac{3}{4}$$

For $$A$$ and $$B$$ to be independent we would need $$P(A \cap B) = P(A)\,P(B)$$. Here

$$P(A)\,P(B) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24}$$

Since $$P(A \cap B) = \dfrac{3}{4} \neq \dfrac{7}{24} = P(A)\,P(B)$$, the events $$A$$ and $$B$$ are not independent.

It is easier to use the complement. 'An odd number at least once' is the opposite of 'no odd number in any of the three throws' (i.e. an even number every time).

On a single throw, the odd numbers are $$\{1,3,5\}$$ and the even numbers are $$\{2,4,6\}$$, so

$$P(\text{odd}) = \frac{3}{6} = \frac{1}{2}, \qquad P(\text{even}) = \frac{3}{6} = \frac{1}{2}$$

The three throws are independent, so the probability of getting an even number on all three throws is

$$P(\text{no odd number}) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$$

Therefore

$$P(\text{odd at least once}) = 1 - P(\text{no odd number}) = 1 - \frac{1}{8} = \frac{7}{8}$$

The only number that is both even and prime is $$2$$ (every other even number is divisible by $$2$$ and hence not prime).

So 'an even prime number on a die' means the die shows a $$2$$. On one die,

$$P(\text{even prime}) = P(2) = \frac{1}{6}$$

The two dice are independent, so the probability of getting an even prime on each die is

$$\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$$

Hence the correct option is (d).

We test each option.

(a) If $$A$$ and $$B$$ are mutually exclusive (and both have non-zero probability), then $$P(A \cap B) = 0$$, whereas independence needs $$P(A \cap B) = P(A)\,P(B) \neq 0$$. So mutual exclusivity does not give independence. ✗

(b) $$P(A'B') = [1 - P(A)][1 - P(B)]$$ means $$P(A' \cap B') = P(A')\,P(B')$$, i.e. $$A'$$ and $$B'$$ are independent. It is a standard result that $$A'$$ and $$B'$$ are independent if and only if $$A$$ and $$B$$ are independent. So this condition is exactly equivalent to the independence of $$A$$ and $$B$$. ✓

(c) $$P(A) = P(B)$$ only says the probabilities are equal — it says nothing about $$P(A \cap B)$$. ✗

(d) $$P(A) + P(B) = 1$$ is unrelated to the product rule for independence. ✗

Hence the correct option is (b).

Let $$A$$ be the event 'there is a strike' and $$E$$ the event 'the job is completed on time'.

Given data:

$$P(A) = 0.65 \quad \Rightarrow \quad P(A') = 1 - 0.65 = 0.35$$

$$P(E|A) = 0.32 \quad \text{(completed on time given a strike)}$$

$$P(E|A') = 0.80 \quad \text{(completed on time given no strike)}$$

The events $$A$$ and $$A'$$ together cover all possibilities, so by the theorem of total probability,

$$P(E) = P(A)\,P(E|A) + P(A')\,P(E|A')$$

$$P(E) = (0.65)(0.32) + (0.35)(0.80)$$

$$P(E) = 0.208 + 0.280 = 0.488$$

Let the events be

$$E_1$$ : Bag I is chosen, $$E_2$$ : Bag II is chosen, $$A$$ : a red ball is drawn.

Since one of the two bags is chosen at random,

$$P(E_1) = P(E_2) = \frac{1}{2}$$

Bag I has $$3$$ red out of $$7$$ balls, and Bag II has $$5$$ red out of $$11$$ balls:

$$P(A|E_1) = \frac{3}{7}, \qquad P(A|E_2) = \frac{5}{11}$$

By Bayes' theorem, the probability that the red ball came from Bag II is

$$P(E_2|A) = \frac{P(E_2)\,P(A|E_2)}{P(E_1)\,P(A|E_1) + P(E_2)\,P(A|E_2)}$$

$$P(E_2|A) = \frac{\dfrac{1}{2} \cdot \dfrac{5}{11}}{\dfrac{1}{2} \cdot \dfrac{3}{7} + \dfrac{1}{2} \cdot \dfrac{5}{11}}$$

Cancel the common factor $$\dfrac{1}{2}$$ from numerator and denominator:

$$P(E_2|A) = \frac{\dfrac{5}{11}}{\dfrac{3}{7} + \dfrac{5}{11}} = \frac{\dfrac{35}{77}}{\dfrac{33}{77} + \dfrac{35}{77}} = \frac{35}{68}$$

Let $$E_1, E_2, E_3$$ be the events that box I, II, III respectively is chosen, and let $$A$$ be the event 'a gold coin is drawn'.

The boxes are identical and chosen at random, so

$$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$$

The probability of drawing a gold coin from each box:

$$P(A|E_1) = 1 \ \text{(both gold)}, \quad P(A|E_2) = 0 \ \text{(both silver)}, \quad P(A|E_3) = \frac{1}{2}$$

The 'other coin' is also gold only if the chosen box is box I (the box with two gold coins). So we need $$P(E_1|A)$$. By Bayes' theorem,

$$P(E_1|A) = \frac{P(E_1)\,P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}$$

$$P(E_1|A) = \frac{\dfrac{1}{3} \cdot 1}{\dfrac{1}{3}\cdot 1 + \dfrac{1}{3}\cdot 0 + \dfrac{1}{3}\cdot \dfrac{1}{2}} = \frac{\dfrac{1}{3}}{\dfrac{1}{3} + 0 + \dfrac{1}{6}}$$

$$P(E_1|A) = \frac{\dfrac{1}{3}}{\dfrac{1}{2}} = \frac{2}{3}$$

Let $$E$$ be the event 'the person has HIV' and $$A$$ the event 'the test reports HIV+ive'.

Since $$0.1\%$$ of the population have HIV,

$$P(E) = \frac{0.1}{100} = 0.001, \qquad P(E') = 1 - 0.001 = 0.999$$

The test detects the disease in $$90\%$$ of HIV-positive people and gives a false positive in $$1\%$$ of HIV-free people:

$$P(A|E) = \frac{90}{100} = 0.9, \qquad P(A|E') = \frac{1}{100} = 0.01$$

By Bayes' theorem,

$$P(E|A) = \frac{P(E)\,P(A|E)}{P(E)\,P(A|E) + P(E')\,P(A|E')}$$

$$P(E|A) = \frac{(0.001)(0.9)}{(0.001)(0.9) + (0.999)(0.01)} = \frac{0.0009}{0.0009 + 0.00999}$$

$$P(E|A) = \frac{0.0009}{0.01089} = \frac{90}{1089} = \frac{10}{121} \approx 0.083$$

So even after a positive report, there is only about an $$8.3\%$$ chance the person actually has HIV.

Let $$E_1, E_2, E_3$$ be the events that a bolt is made by machine A, B, C respectively, and let $$D$$ be the event 'the bolt is defective'.

From the production shares,

$$P(E_1) = 0.25, \qquad P(E_2) = 0.35, \qquad P(E_3) = 0.40$$

From the defective rates,

$$P(D|E_1) = 0.05, \qquad P(D|E_2) = 0.04, \qquad P(D|E_3) = 0.02$$

By Bayes' theorem, the probability the defective bolt was made by machine B is

$$P(E_2|D) = \frac{P(E_2)\,P(D|E_2)}{P(E_1)P(D|E_1) + P(E_2)P(D|E_2) + P(E_3)P(D|E_3)}$$

Compute each term:

$$P(E_1)P(D|E_1) = 0.25 \times 0.05 = 0.0125$$

$$P(E_2)P(D|E_2) = 0.35 \times 0.04 = 0.0140$$

$$P(E_3)P(D|E_3) = 0.40 \times 0.02 = 0.0080$$

Total $$= 0.0125 + 0.0140 + 0.0080 = 0.0345$$. Hence

$$P(E_2|D) = \frac{0.0140}{0.0345} = \frac{140}{345} = \frac{28}{69}$$

Let $$T_1, T_2, T_3, T_4$$ be the events that the doctor comes by train, bus, scooter and other means respectively, and let $$L$$ be the event 'he is late'.

$$P(T_1) = \frac{3}{10}, \quad P(T_2) = \frac{1}{5}, \quad P(T_3) = \frac{1}{10}, \quad P(T_4) = \frac{2}{5}$$

$$P(L|T_1) = \frac{1}{4}, \quad P(L|T_2) = \frac{1}{3}, \quad P(L|T_3) = \frac{1}{12}, \quad P(L|T_4) = 0$$

By Bayes' theorem,

$$P(T_1|L) = \frac{P(T_1)\,P(L|T_1)}{P(T_1)P(L|T_1) + P(T_2)P(L|T_2) + P(T_3)P(L|T_3) + P(T_4)P(L|T_4)}$$

Compute each term:

$$P(T_1)P(L|T_1) = \frac{3}{10} \times \frac{1}{4} = \frac{3}{40} = \frac{9}{120}$$

$$P(T_2)P(L|T_2) = \frac{1}{5} \times \frac{1}{3} = \frac{1}{15} = \frac{8}{120}$$

$$P(T_3)P(L|T_3) = \frac{1}{10} \times \frac{1}{12} = \frac{1}{120}$$

$$P(T_4)P(L|T_4) = \frac{2}{5} \times 0 = 0$$

Total $$= \dfrac{9 + 8 + 1 + 0}{120} = \dfrac{18}{120}$$. Hence

$$P(T_1|L) = \frac{\;9/120\;}{\;18/120\;} = \frac{9}{18} = \frac{1}{2}$$

Let $$E$$ be the event 'the die actually shows a six' and $$A$$ the event 'the man reports a six'.

For a fair die,

$$P(E) = \frac{1}{6}, \qquad P(E') = \frac{5}{6}$$

If a six really came up $$(E)$$, the man reports a six only when he speaks the truth:

$$P(A|E) = \frac{3}{4}$$

If a six did not come up $$(E')$$, the man reports a six only when he lies:

$$P(A|E') = 1 - \frac{3}{4} = \frac{1}{4}$$

By Bayes' theorem,

$$P(E|A) = \frac{P(E)\,P(A|E)}{P(E)\,P(A|E) + P(E')\,P(A|E')}$$

$$P(E|A) = \frac{\dfrac{1}{6} \cdot \dfrac{3}{4}}{\dfrac{1}{6} \cdot \dfrac{3}{4} + \dfrac{5}{6} \cdot \dfrac{1}{4}} = \frac{\dfrac{3}{24}}{\dfrac{3}{24} + \dfrac{5}{24}} = \frac{3}{8}$$

Initially the urn has $$5$$ red and $$5$$ black balls ($$10$$ in all). The colour of the first ball decides what the urn contains for the second draw, so we split into two cases.

Case 1 — first ball is red.

$$P(\text{1st red}) = \frac{5}{10} = \frac{1}{2}$$

The red ball is returned and $$2$$ more red balls added, so the urn now has $$7$$ red and $$5$$ black ($$12$$ balls):

$$P(\text{2nd red} \mid \text{1st red}) = \frac{7}{12}$$

Case 2 — first ball is black.

$$P(\text{1st black}) = \frac{5}{10} = \frac{1}{2}$$

The black ball is returned and $$2$$ more black balls added, so the urn now has $$5$$ red and $$7$$ black ($$12$$ balls):

$$P(\text{2nd red} \mid \text{1st black}) = \frac{5}{12}$$

By the theorem of total probability,

$$P(\text{2nd ball red}) = \frac{1}{2} \cdot \frac{7}{12} + \frac{1}{2} \cdot \frac{5}{12} = \frac{7}{24} + \frac{5}{24} = \frac{12}{24} = \frac{1}{2}$$

Let $$E_1$$ : the first bag is chosen, $$E_2$$ : the second bag is chosen, and $$A$$ : the drawn ball is red.

One of the two bags is chosen at random, so

$$P(E_1) = P(E_2) = \frac{1}{2}$$

Bag 1 has $$4$$ red out of $$8$$ balls; bag 2 has $$2$$ red out of $$8$$ balls:

$$P(A|E_1) = \frac{4}{8} = \frac{1}{2}, \qquad P(A|E_2) = \frac{2}{8} = \frac{1}{4}$$

By Bayes' theorem,

$$P(E_1|A) = \frac{P(E_1)\,P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$$

$$P(E_1|A) = \frac{\dfrac{1}{2} \cdot \dfrac{1}{2}}{\dfrac{1}{2} \cdot \dfrac{1}{2} + \dfrac{1}{2} \cdot \dfrac{1}{4}} = \frac{\dfrac{1}{4}}{\dfrac{1}{4} + \dfrac{1}{8}} = \frac{\dfrac{2}{8}}{\dfrac{3}{8}} = \frac{2}{3}$$

Let $$E_1$$ : the student is a hosteller, $$E_2$$ : the student is a day scholar, and $$A$$ : the student has an A grade.

From the given percentages,

$$P(E_1) = 0.60, \qquad P(E_2) = 0.40$$

$$P(A|E_1) = 0.30, \qquad P(A|E_2) = 0.20$$

By Bayes' theorem,

$$P(E_1|A) = \frac{P(E_1)\,P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$$

$$P(E_1|A) = \frac{(0.60)(0.30)}{(0.60)(0.30) + (0.40)(0.20)} = \frac{0.18}{0.18 + 0.08} = \frac{0.18}{0.26}$$

$$P(E_1|A) = \frac{18}{26} = \frac{9}{13}$$

Let $$E_1$$ : the student knows the answer, $$E_2$$ : the student guesses, and $$A$$ : the answer is correct.

$$P(E_1) = \frac{3}{4}, \qquad P(E_2) = \frac{1}{4}$$

If the student knows the answer, it is certainly correct; if he guesses, it is correct with probability $$\frac{1}{4}$$:

$$P(A|E_1) = 1, \qquad P(A|E_2) = \frac{1}{4}$$

By Bayes' theorem,

$$P(E_1|A) = \frac{P(E_1)\,P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$$

$$P(E_1|A) = \frac{\dfrac{3}{4} \cdot 1}{\dfrac{3}{4} \cdot 1 + \dfrac{1}{4} \cdot \dfrac{1}{4}} = \frac{\dfrac{3}{4}}{\dfrac{3}{4} + \dfrac{1}{16}}$$

$$P(E_1|A) = \frac{\dfrac{12}{16}}{\dfrac{12}{16} + \dfrac{1}{16}} = \frac{12}{13}$$

Let $$E_1$$ : the person has the disease, $$E_2$$ : the person is healthy, and $$A$$ : the test result is positive.

Since $$0.1\%$$ of the population has the disease,

$$P(E_1) = \frac{0.1}{100} = 0.001, \qquad P(E_2) = 1 - 0.001 = 0.999$$

The test detects the disease with probability $$0.99$$ when present, and gives a false positive with probability $$0.005$$ for a healthy person:

$$P(A|E_1) = 0.99, \qquad P(A|E_2) = 0.005$$

By Bayes' theorem,

$$P(E_1|A) = \frac{P(E_1)\,P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$$

$$P(E_1|A) = \frac{(0.001)(0.99)}{(0.001)(0.99) + (0.999)(0.005)} = \frac{0.00099}{0.00099 + 0.004995}$$

$$P(E_1|A) = \frac{0.00099}{0.005985} = \frac{990}{5985} = \frac{22}{133} \approx 0.165$$

Let $$E_1$$ : the two-headed coin is chosen, $$E_2$$ : the biased coin is chosen, $$E_3$$ : the unbiased coin is chosen, and $$A$$ : the toss shows heads.

One of the three coins is chosen at random, so

$$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$$

The probability of heads for each coin:

$$P(A|E_1) = 1 \ \text{(both faces heads)}, \quad P(A|E_2) = \frac{75}{100} = \frac{3}{4}, \quad P(A|E_3) = \frac{1}{2}$$

By Bayes' theorem,

$$P(E_1|A) = \frac{P(E_1)\,P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}$$

$$P(E_1|A) = \frac{\dfrac{1}{3} \cdot 1}{\dfrac{1}{3}\cdot 1 + \dfrac{1}{3}\cdot \dfrac{3}{4} + \dfrac{1}{3}\cdot \dfrac{1}{2}}$$

Cancelling the common factor $$\dfrac{1}{3}$$:

$$P(E_1|A) = \frac{1}{1 + \dfrac{3}{4} + \dfrac{1}{2}} = \frac{1}{\dfrac{4 + 3 + 2}{4}} = \frac{1}{\dfrac{9}{4}} = \frac{4}{9}$$

The total number of insured persons is $$2000 + 4000 + 6000 = 12000$$.

Let $$E_1$$ : scooter driver, $$E_2$$ : car driver, $$E_3$$ : truck driver, and $$A$$ : the person meets with an accident.

$$P(E_1) = \frac{2000}{12000} = \frac{1}{6}, \quad P(E_2) = \frac{4000}{12000} = \frac{1}{3}, \quad P(E_3) = \frac{6000}{12000} = \frac{1}{2}$$

$$P(A|E_1) = 0.01, \qquad P(A|E_2) = 0.03, \qquad P(A|E_3) = 0.15$$

By Bayes' theorem,

$$P(E_1|A) = \frac{P(E_1)\,P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}$$

Compute each term (writing all over $$600$$):

$$P(E_1)P(A|E_1) = \frac{1}{6} \times 0.01 = \frac{1}{600}$$

$$P(E_2)P(A|E_2) = \frac{1}{3} \times 0.03 = 0.01 = \frac{6}{600}$$

$$P(E_3)P(A|E_3) = \frac{1}{2} \times 0.15 = 0.075 = \frac{45}{600}$$

Total $$= \dfrac{1 + 6 + 45}{600} = \dfrac{52}{600}$$. Hence

$$P(E_1|A) = \frac{\;1/600\;}{\;52/600\;} = \frac{1}{52}$$

Let $$E_1$$ : the item is produced by machine A, $$E_2$$ : the item is produced by machine B, and $$D$$ : the item is defective.

$$P(E_1) = 0.60, \qquad P(E_2) = 0.40$$

$$P(D|E_1) = 0.02, \qquad P(D|E_2) = 0.01$$

By Bayes' theorem,

$$P(E_2|D) = \frac{P(E_2)\,P(D|E_2)}{P(E_1)P(D|E_1) + P(E_2)P(D|E_2)}$$

$$P(E_2|D) = \frac{(0.40)(0.01)}{(0.60)(0.02) + (0.40)(0.01)} = \frac{0.004}{0.012 + 0.004} = \frac{0.004}{0.016}$$

$$P(E_2|D) = \frac{4}{16} = \frac{1}{4}$$

Let $$E_1$$ : the first group wins, $$E_2$$ : the second group wins, and $$A$$ : a new product is introduced.

$$P(E_1) = 0.6, \qquad P(E_2) = 0.4$$

$$P(A|E_1) = 0.7, \qquad P(A|E_2) = 0.3$$

We want the probability that the second group was responsible, given that a new product was introduced. By Bayes' theorem,

$$P(E_2|A) = \frac{P(E_2)\,P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$$

$$P(E_2|A) = \frac{(0.4)(0.3)}{(0.6)(0.7) + (0.4)(0.3)} = \frac{0.12}{0.42 + 0.12} = \frac{0.12}{0.54}$$

$$P(E_2|A) = \frac{12}{54} = \frac{2}{9}$$

Let $$E_1$$ : the die shows $$5$$ or $$6$$, $$E_2$$ : the die shows $$1, 2, 3$$ or $$4$$, and $$A$$ : exactly one head is obtained.

$$P(E_1) = \frac{2}{6} = \frac{1}{3}, \qquad P(E_2) = \frac{4}{6} = \frac{2}{3}$$

Case $$E_1$$ (coin tossed three times): exactly one head out of three tosses can occur as $$HTT, THT, TTH$$:

$$P(A|E_1) = \frac{3}{8}$$

Case $$E_2$$ (coin tossed once): exactly one head means the single toss is a head:

$$P(A|E_2) = \frac{1}{2}$$

By Bayes' theorem,

$$P(E_2|A) = \frac{P(E_2)\,P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$$

$$P(E_2|A) = \frac{\dfrac{2}{3} \cdot \dfrac{1}{2}}{\dfrac{1}{3} \cdot \dfrac{3}{8} + \dfrac{2}{3} \cdot \dfrac{1}{2}} = \frac{\dfrac{1}{3}}{\dfrac{1}{8} + \dfrac{1}{3}}$$

$$P(E_2|A) = \frac{\dfrac{1}{3}}{\dfrac{3}{24} + \dfrac{8}{24}} = \frac{\dfrac{1}{3}}{\dfrac{11}{24}} = \frac{1}{3} \times \frac{24}{11} = \frac{8}{11}$$

Let $$E_1, E_2, E_3$$ be the events that operator A, B, C respectively is on the job, and let $$D$$ be the event 'the item produced is defective'.

$$P(E_1) = 0.50, \qquad P(E_2) = 0.30, \qquad P(E_3) = 0.20$$

$$P(D|E_1) = 0.01, \qquad P(D|E_2) = 0.05, \qquad P(D|E_3) = 0.07$$

By Bayes' theorem,

$$P(E_1|D) = \frac{P(E_1)\,P(D|E_1)}{P(E_1)P(D|E_1) + P(E_2)P(D|E_2) + P(E_3)P(D|E_3)}$$

Compute each term:

$$P(E_1)P(D|E_1) = 0.50 \times 0.01 = 0.005$$

$$P(E_2)P(D|E_2) = 0.30 \times 0.05 = 0.015$$

$$P(E_3)P(D|E_3) = 0.20 \times 0.07 = 0.014$$

Total $$= 0.005 + 0.015 + 0.014 = 0.034$$. Hence

$$P(E_1|D) = \frac{0.005}{0.034} = \frac{5}{34}$$

Let $$E_1$$ : the lost card is a diamond, $$E_2$$ : the lost card is not a diamond, and $$A$$ : the two cards drawn from the remaining $$51$$ are both diamonds.

A pack has $$13$$ diamonds out of $$52$$ cards, so

$$P(E_1) = \frac{13}{52} = \frac{1}{4}, \qquad P(E_2) = \frac{39}{52} = \frac{3}{4}$$

If the lost card is a diamond, the remaining $$51$$ cards contain $$12$$ diamonds:

$$P(A|E_1) = \frac{\binom{12}{2}}{\binom{51}{2}} = \frac{12 \times 11}{51 \times 50}$$

If the lost card is not a diamond, the remaining $$51$$ cards still contain all $$13$$ diamonds:

$$P(A|E_2) = \frac{\binom{13}{2}}{\binom{51}{2}} = \frac{13 \times 12}{51 \times 50}$$

By Bayes' theorem,

$$P(E_1|A) = \frac{P(E_1)\,P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$$

$$P(E_1|A) = \frac{\dfrac{1}{4} \cdot \dfrac{12 \times 11}{51 \times 50}}{\dfrac{1}{4} \cdot \dfrac{12 \times 11}{51 \times 50} + \dfrac{3}{4} \cdot \dfrac{13 \times 12}{51 \times 50}}$$

Cancel the common factor $$\dfrac{1}{4} \cdot \dfrac{12}{51 \times 50}$$ from numerator and denominator:

$$P(E_1|A) = \frac{11}{11 + 3 \times 13} = \frac{11}{11 + 39} = \frac{11}{50}$$

Let $$E_1$$ : a head actually appears, $$E_2$$ : a tail actually appears, and $$R$$ : A reports a head.

For a fair coin,

$$P(E_1) = \frac{1}{2}, \qquad P(E_2) = \frac{1}{2}$$

If a head appeared, A reports a head by speaking the truth; if a tail appeared, A reports a head by lying:

$$P(R|E_1) = \frac{4}{5}, \qquad P(R|E_2) = 1 - \frac{4}{5} = \frac{1}{5}$$

By Bayes' theorem,

$$P(E_1|R) = \frac{P(E_1)\,P(R|E_1)}{P(E_1)P(R|E_1) + P(E_2)P(R|E_2)}$$

$$P(E_1|R) = \frac{\dfrac{1}{2} \cdot \dfrac{4}{5}}{\dfrac{1}{2} \cdot \dfrac{4}{5} + \dfrac{1}{2} \cdot \dfrac{1}{5}} = \frac{\dfrac{4}{10}}{\dfrac{4}{10} + \dfrac{1}{10}} = \frac{4}{5}$$

Hence the correct option is (a).

Step 1: Use the subset condition.

Since $$A \subset B$$, every outcome of $$A$$ also lies in $$B$$, so

$$A \cap B = A.$$

Step 2: Apply the definition of conditional probability.

Because $$P(B) \neq 0$$,

$$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)}.$$

Step 3: Compare $$P(A \mid B)$$ with $$P(A)$$.

Since $$P(B)$$ is a probability, $$0 \lt P(B) \leq 1$$. Taking reciprocals reverses the inequality:

$$\frac{1}{P(B)} \geq 1.$$

Multiplying the non-negative quantity $$P(A)$$ by a factor that is at least $$1$$ cannot decrease it, so

$$P(A \mid B) = P(A) \cdot \frac{1}{P(B)} \geq P(A).$$

Step 4: Identify the correct option.

We have shown $$P(A \mid B) \geq P(A)$$. Option (a) is wrong because the ratio is inverted (the correct ratio is $$P(A)/P(B)$$, not $$P(B)/P(A)$$), and option (b) contradicts the inequality just derived. Therefore the correct choice is (c).

Step 1 Define the events

Let

• $$E_1, E_2, E_3, E_4$$ be the events that the selected box is I, II, III and IV respectively,
• $$B$$ be the event that the ball drawn is black.

A box is chosen at random, so

$$P(E_1)=P(E_2)=P(E_3)=P(E_4)=\frac14.$$

Step 2 Conditional probability of drawing a black ball from each box

Step 3 Total probability of drawing a black ball

$$P(B)=\sum_{i=1}^{4}P(E_i)\,P(B\mid E_i)=\frac14\left(\frac16+\frac14+\frac17+\frac{4}{13}\right).$$

Convert to the common denominator 1092:

$$\frac16=\frac{182}{1092},\;\frac14=\frac{273}{1092},\;\frac17=\frac{156}{1092},\;\frac{4}{13}=\frac{336}{1092}.$$

Hence

$$P(B)=\frac14\cdot\frac{182+273+156+336}{1092}=\frac14\cdot\frac{947}{1092}=\frac{947}{4368}.$$

Step 4 Bayes' theorem for the required probability

$$P(E_3\mid B)=\frac{P(E_3)\,P(B\mid E_3)}{P(B)}=\frac{\dfrac14\times\dfrac17}{\dfrac{947}{4368}}=\frac1{28}\times\frac{4368}{947}=\frac{156}{947}.$$

Conclusion

The probability that the black ball was drawn from box III is $$\boxed{\dfrac{156}{947}\approx0.165}$$.

On any throw, getting a six and not getting a six have probabilities

$$P(6) = \frac{1}{6}, \qquad P(\text{not } 6) = \frac{5}{6}$$

A throws first, so A can win on the 1st, 3rd, 5th, ... throw of the game. A wins on his $$k$$-th turn only if all preceding throws (by A and B) failed to be a six.

$$P(\text{A wins on 1st throw}) = \frac{1}{6}$$

$$P(\text{A wins on 3rd throw}) = \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} = \left(\frac{5}{6}\right)^2 \frac{1}{6}$$

$$P(\text{A wins on 5th throw}) = \left(\frac{5}{6}\right)^4 \frac{1}{6}, \quad \text{and so on.}$$

Adding all these mutually exclusive possibilities gives an infinite geometric series with first term $$a = \dfrac{1}{6}$$ and common ratio $$r = \left(\dfrac{5}{6}\right)^2 = \dfrac{25}{36}$$:

$$P(\text{A wins}) = \frac{1}{6}\left[1 + \left(\frac{5}{6}\right)^2 + \left(\frac{5}{6}\right)^4 + \cdots\right] = \frac{1}{6} \cdot \frac{1}{1 - \dfrac{25}{36}}$$

$$P(\text{A wins}) = \frac{1}{6} \cdot \frac{36}{11} = \frac{6}{11}$$

Since one of them must eventually win,

$$P(\text{B wins}) = 1 - P(\text{A wins}) = 1 - \frac{6}{11} = \frac{5}{11}$$

Let $$C$$ : the machine is correctly set up, $$I$$ : the machine is incorrectly set up, and $$A$$ : the machine produces $$2$$ acceptable items.

From past experience,

$$P(C) = 0.8, \qquad P(I) = 0.2$$

If the setup is correct, each item is acceptable with probability $$0.9$$; if incorrect, with probability $$0.4$$. Treating the two items independently,

$$P(A|C) = 0.9 \times 0.9 = 0.81$$

$$P(A|I) = 0.4 \times 0.4 = 0.16$$

By Bayes' theorem,

$$P(C|A) = \frac{P(C)\,P(A|C)}{P(C)P(A|C) + P(I)P(A|I)}$$

$$P(C|A) = \frac{(0.8)(0.81)}{(0.8)(0.81) + (0.2)(0.16)} = \frac{0.648}{0.648 + 0.032} = \frac{0.648}{0.680}$$

$$P(C|A) = \frac{648}{680} = \frac{81}{85}$$

Let $$E_1$$ : the person is male, $$E_2$$ : the person is female, and $$G$$ : the person has grey hair.

There are equal numbers of males and females, so

$$P(E_1) = P(E_2) = \frac{1}{2}$$

$$5\%$$ of men and $$0.25\%$$ of women have grey hair:

$$P(G|E_1) = \frac{5}{100} = 0.05, \qquad P(G|E_2) = \frac{0.25}{100} = 0.0025$$

By Bayes' theorem,

$$P(E_1|G) = \frac{P(E_1)\,P(G|E_1)}{P(E_1)P(G|E_1) + P(E_2)P(G|E_2)}$$

$$P(E_1|G) = \frac{\dfrac{1}{2}(0.05)}{\dfrac{1}{2}(0.05) + \dfrac{1}{2}(0.0025)} = \frac{0.05}{0.05 + 0.0025} = \frac{0.05}{0.0525}$$

$$P(E_1|G) = \frac{500}{525} = \frac{20}{21}$$

Whether a person is right-handed is a Bernoulli trial. Let $$X$$ be the number of right-handed people among the $$n = 10$$ sampled. Then $$X$$ follows a binomial distribution with

$$p = P(\text{right-handed}) = \frac{90}{100} = \frac{9}{10}, \qquad q = 1 - p = \frac{1}{10}$$

$$P(X = r) = \binom{10}{r} p^{r} q^{\,10 - r} = \binom{10}{r}\left(\frac{9}{10}\right)^{r}\left(\frac{1}{10}\right)^{10 - r}$$

We need $$P(X \le 6)$$, the probability that at most $$6$$ are right-handed. It is easier to use the complement — 'at most 6' is the opposite of 'at least 7':

$$P(X \le 6) = 1 - P(X \ge 7) = 1 - \big[P(7) + P(8) + P(9) + P(10)\big]$$

$$P(X \le 6) = 1 - \sum_{r=7}^{10} \binom{10}{r}\left(\frac{9}{10}\right)^{r}\left(\frac{1}{10}\right)^{10 - r}$$

Evaluating the four terms:

$$P(7) = 120\,(0.9)^7(0.1)^3 \approx 0.0574$$

$$P(8) = 45\,(0.9)^8(0.1)^2 \approx 0.1937$$

$$P(9) = 10\,(0.9)^9(0.1)^1 \approx 0.3874$$

$$P(10) = (0.9)^{10} \approx 0.3487$$

So $$P(X \ge 7) \approx 0.9872$$, and

$$P(X \le 6) \approx 1 - 0.9872 = 0.0128$$

A leap year has $$366$$ days. Dividing by $$7$$:

$$366 = 7 \times 52 + 2$$

So a leap year is made up of $$52$$ complete weeks plus $$2$$ extra days. The $$52$$ complete weeks already contain exactly $$52$$ Tuesdays. To get a 53rd Tuesday, at least one of the $$2$$ extra days must be a Tuesday.

The $$2$$ extra days are a pair of consecutive days. The equally likely possibilities are:

$$\{\text{Sun, Mon}\},\ \{\text{Mon, Tue}\},\ \{\text{Tue, Wed}\},\ \{\text{Wed, Thu}\},\ \{\text{Thu, Fri}\},\ \{\text{Fri, Sat}\},\ \{\text{Sat, Sun}\}$$

This is a sample space of $$7$$ equally likely outcomes. A Tuesday appears among the extra days in exactly $$2$$ of them: $$\{\text{Mon, Tue}\}$$ and $$\{\text{Tue, Wed}\}$$.

Therefore

$$P(53 \text{ Tuesdays}) = \frac{2}{7}$$

Let $$E_A, E_B, E_C, E_D$$ be the events that box A, B, C, D respectively is selected, and let $$R$$ be the event 'a red marble is drawn'.

One box is selected at random, so

$$P(E_A) = P(E_B) = P(E_C) = P(E_D) = \frac{1}{4}$$

Each box contains $$10$$ marbles. The probability of drawing a red marble from each box:

$$P(R|E_A) = \frac{1}{10}, \quad P(R|E_B) = \frac{6}{10}, \quad P(R|E_C) = \frac{8}{10}, \quad P(R|E_D) = \frac{0}{10} = 0$$

Total probability of a red marble:

$$P(R) = \frac{1}{4}\left(\frac{1}{10} + \frac{6}{10} + \frac{8}{10} + 0\right) = \frac{1}{4} \times \frac{15}{10} = \frac{15}{40}$$

By Bayes' theorem, $$P(E_X|R) = \dfrac{P(E_X)\,P(R|E_X)}{P(R)}$$:

Box A: $$\quad P(E_A|R) = \dfrac{\dfrac{1}{4}\cdot\dfrac{1}{10}}{\dfrac{15}{40}} = \dfrac{1/40}{15/40} = \dfrac{1}{15}$$

Box B: $$\quad P(E_B|R) = \dfrac{\dfrac{1}{4}\cdot\dfrac{6}{10}}{\dfrac{15}{40}} = \dfrac{6/40}{15/40} = \dfrac{6}{15} = \dfrac{2}{5}$$

Box C: $$\quad P(E_C|R) = \dfrac{\dfrac{1}{4}\cdot\dfrac{8}{10}}{\dfrac{15}{40}} = \dfrac{8/40}{15/40} = \dfrac{8}{15}$$

Let $$E_1$$ : the patient follows the meditation and yoga course, $$E_2$$ : the patient takes the drug, and $$A$$ : the patient suffers a heart attack.

The patient chooses either option with equal probability:

$$P(E_1) = P(E_2) = \frac{1}{2}$$

The base chance of a heart attack is $$40\% = 0.40$$.

Meditation and yoga reduce this risk by $$30\%$$, so the remaining risk is

$$P(A|E_1) = 0.40 \times (1 - 0.30) = 0.40 \times 0.70 = 0.28$$

The drug reduces the risk by $$25\%$$, so

$$P(A|E_2) = 0.40 \times (1 - 0.25) = 0.40 \times 0.75 = 0.30$$

By Bayes' theorem,

$$P(E_1|A) = \frac{P(E_1)\,P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$$

$$P(E_1|A) = \frac{\dfrac{1}{2}(0.28)}{\dfrac{1}{2}(0.28) + \dfrac{1}{2}(0.30)} = \frac{0.28}{0.28 + 0.30} = \frac{0.28}{0.58}$$

$$P(E_1|A) = \frac{28}{58} = \frac{14}{29}$$

A second order determinant has $$4$$ entries, each chosen as $$0$$ or $$1$$. Write it as

$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$$

Each entry is $$0$$ or $$1$$ independently, so the total number of equally likely determinants is

$$2 \times 2 \times 2 \times 2 = 16$$

Since $$a, b, c, d \in \{0, 1\}$$, both products $$ad$$ and $$bc$$ lie in $$\{0, 1\}$$, so the value $$ad - bc$$ can only be $$-1$$, $$0$$ or $$1$$. The determinant is positive only when

$$ad - bc = 1 \quad \Longleftrightarrow \quad ad = 1 \ \text{and} \ bc = 0$$

$$ad = 1$$: requires $$a = 1$$ and $$d = 1$$ — only $$1$$ way.

$$bc = 0$$: requires $$b$$ and $$c$$ not both equal to $$1$$. Of the $$4$$ combinations of $$(b, c)$$, only $$(1,1)$$ fails, so there are $$3$$ ways.

Number of favourable determinants $$= 1 \times 3 = 3$$.

$$P(\text{determinant positive}) = \frac{3}{16}$$

Let $$E_1$$ : the transferred ball is red, $$E_2$$ : the transferred ball is black, and $$A$$ : the ball later drawn from Bag II is red.

Bag I has $$3$$ red and $$4$$ black ($$7$$ balls), so the transferred ball is

$$P(E_1) = \frac{3}{7}, \qquad P(E_2) = \frac{4}{7}$$

Bag II originally has $$4$$ red and $$5$$ black ($$9$$ balls); after the transfer it has $$10$$ balls.

If a red ball was transferred $$(E_1)$$: Bag II has $$5$$ red and $$5$$ black, so

$$P(A|E_1) = \frac{5}{10} = \frac{1}{2}$$

If a black ball was transferred $$(E_2)$$: Bag II has $$4$$ red and $$6$$ black, so

$$P(A|E_2) = \frac{4}{10} = \frac{2}{5}$$

By Bayes' theorem,

$$P(E_2|A) = \frac{P(E_2)\,P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$$

$$P(E_2|A) = \frac{\dfrac{4}{7} \cdot \dfrac{2}{5}}{\dfrac{3}{7} \cdot \dfrac{1}{2} + \dfrac{4}{7} \cdot \dfrac{2}{5}} = \frac{\dfrac{8}{35}}{\dfrac{3}{14} + \dfrac{8}{35}}$$

Writing the denominator over the LCM $$70$$: $$\dfrac{3}{14} = \dfrac{15}{70}$$ and $$\dfrac{8}{35} = \dfrac{16}{70}$$, so the sum is $$\dfrac{31}{70}$$. Also $$\dfrac{8}{35} = \dfrac{16}{70}$$. Hence

$$P(E_2|A) = \frac{\;16/70\;}{\;31/70\;} = \frac{16}{31}$$

Use the definition of conditional probability (valid since $$P(A) \neq 0$$):

$$P(B|A) = \frac{P(A \cap B)}{P(A)}$$

We are given $$P(B|A) = 1$$, so

$$\frac{P(A \cap B)}{P(A)} = 1 \quad \Rightarrow \quad P(A \cap B) = P(A)$$

Now $$A \cap B$$ is always a subset of $$A$$, and here it has the same probability as $$A$$. This means no part of $$A$$ lies outside $$B$$ — every outcome of $$A$$ is also an outcome of $$B$$. Hence

$$A \subset B$$

So the correct option is (a).

Start from the given inequality and write $$P(A|B)$$ using its definition:

$$P(A|B) > P(A) \quad \Rightarrow \quad \frac{P(A \cap B)}{P(B)} > P(A)$$

Multiplying both sides by $$P(B) > 0$$:

$$P(A \cap B) > P(A)\,P(B) \qquad \text{...(1)}$$

(So option (b), which claims the reverse inequality, is wrong.)

Now examine $$P(B|A)$$. Dividing (1) by $$P(A) > 0$$:

$$\frac{P(A \cap B)}{P(A)} > P(B) \quad \Rightarrow \quad P(B|A) > P(B)$$

Hence the correct option is (c).

The given condition is

$$P(A) + P(B) - P(A \cap B) = P(A)$$

Cancel $$P(A)$$ from both sides:

$$P(B) - P(A \cap B) = 0 \quad \Rightarrow \quad P(A \cap B) = P(B)$$

Now compute $$P(A|B)$$ using its definition (assuming $$P(B) \neq 0$$):

$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(B)}{P(B)} = 1$$

Hence the correct option is (b).