Chapter 13: Probability

Step 1 : Define the sample space
When a fair coin is tossed once, only two distinct outcomes are possible—Head (H) or Tail (T). Hence the sample space is
$$S = \{\mathrm{H}, \mathrm{T}\}$$
Therefore
$$n(S) = 2$$

Step 2 : Probability of getting a head
Let the event of getting a head be denoted by $$E_H$$. Then
$$E_H = \{\mathrm{H}\} \quad \Rightarrow \quad n(E_H) = 1$$
Using the classical definition of probability,
$$P(\text{head}) = P(E_H) = \frac{n(E_H)}{n(S)} = \frac{1}{2}$$

Step 3 : Probability of getting a tail
Let the event of getting a tail be denoted by $$E_T$$. Then
$$E_T = \{\mathrm{T}\} \quad \Rightarrow \quad n(E_T) = 1$$
Thus
$$P(\text{tail}) = P(E_T) = \frac{n(E_T)}{n(S)} = \frac{1}{2}$$

Step 4 : Verification
For any experiment, the probabilities of all mutually exclusive and exhaustive outcomes must add up to 1. Indeed,
$$P(\text{head}) + P(\text{tail}) = \frac{1}{2} + \frac{1}{2} = 1$$
which is consistent with theory.

Therefore, the probability of getting either a head or a tail on a single toss of a fair coin is \(\frac12\) in each case.

Step 1: Sample space
A standard die has six faces, so the possible outcomes when it is thrown once are
$$S = \{1,2,3,4,5,6\}$$
Thus, $$n(S) = 6$$.

Step 2: Events of interest

• (i) Getting a number greater than $$4$$:
  $$A = \{5,6\} \;\Rightarrow\; n(A)=2$$
• (ii) Getting a number less than or equal to $$4$$:
  $$B = \{1,2,3,4\} \;\Rightarrow\; n(B)=4$$

Step 3: Probability formula
For any event $$E$$, $$P(E)=\dfrac{n(E)}{n(S)}$$.

(i) $$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{2}{6}=\dfrac{1}{3}$$.

(ii) $$P(B)=\dfrac{n(B)}{n(S)}=\dfrac{4}{6}=\dfrac{2}{3}$$.

Check
Events $$A$$ and $$B$$ are complementary, hence
$$P(A)+P(B)=\dfrac{1}{3}+\dfrac{2}{3}=1$$, confirming the computations.

We are given the probability that Sangeeta wins the tennis match.

Let

• $$P(S)$$ = probability that Sangeeta wins.
• $$P(R)$$ = probability that Reshma wins.

The question states that

$$P(S) = 0.62$$

In a tennis match there are no ties: exactly one of the two players must win. Hence the two outcomes "Sangeeta wins" and "Reshma wins" are exhaustive and mutually exclusive.

Therefore, by the fundamental rule for the probabilities of complementary events,

$$P(S) + P(R) = 1$$

We substitute the known value and solve for $$P(R)$$:

$$0.62 + P(R) = 1$$

Subtract $$0.62$$ from both sides:

$$P(R) = 1 - 0.62$$

Carry out the subtraction:

$$P(R) = 0.38$$

Thus, the probability that Reshma wins the match is $$0.38$$.

Let the possible birthdays be the 365 different calendar days of a non-leap year.

Total number of equally likely outcomes

For each of the two friends, the birthday can be chosen in $$365$$ ways. Using the Fundamental Principle of Counting (FPC), the number of ordered pairs of birthdays is therefore

$$365 \times 365 = 365^2.$$

(i) Probability that the birthdays are different

• Savita can be born on any of the $$365$$ days.
• To be different, Hamida must then be born on any of the remaining $$365 - 1 = 364$$ days.

Hence, the number of favourable outcomes is $$365 \times 364.$$

Probability

$$P(\text{different birthdays}) = \dfrac{365 \times 364}{365^2} = \dfrac{364}{365}.$$

(ii) Probability that the birthdays are the same

• The common birthday can be any one of the $$365$$ days.
• Once that day is fixed, both friends must have that very day as their birthday. This yields exactly $$365$$ favourable ordered pairs.

Thus,

$$P(\text{same birthday}) = \dfrac{365}{365^2} = \dfrac{1}{365}.$$

As a check, $$\dfrac{364}{365} + \dfrac{1}{365} = 1,$$ confirming that the two complementary events together exhaust all possibilities.

Given data

• Total number of students in the class = $$40$$
• Number of girls = $$25$$
• Number of boys = $$15$$

Each student’s name is written on one identical card and one card is drawn at random. Every card is therefore equally likely to be selected.

Theoretical probability of an event $$E$$ is

$$\text{P}(E)=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$$

Total number of possible outcomes = total cards = $$40$$.

(i) Event E₁: “the card bears the name of a girl”

• Favourable outcomes = number of girls = $$25$$

Probability

$$\text{P}(E_1)=\dfrac{25}{40}=\dfrac{5}{8}$$

(ii) Event E₂: “the card bears the name of a boy”

• Favourable outcomes = number of boys = $$15$$

Probability

$$\text{P}(E_2)=\dfrac{15}{40}=\dfrac{3}{8}$$

Check: $$\text{P}(E_1)+\text{P}(E_2)=\dfrac{5}{8}+\dfrac{3}{8}=1$$, confirming the probabilities are consistent.

When two different coins (₹1-coin and ₹2-coin) are tossed together, every coin can show either Head (H) or Tail (T).

Step 1 – List the sample space
For each of the two coins write the first letter (H/T), keeping the order ₹1-coin first, ₹2-coin second:

$$S = \{\text{HH},\;\text{HT},\;\text{TH},\;\text{TT}\}$$

Hence, the total number of equally likely outcomes is $$n(S)=4$$.

Step 2 – Identify the favourable outcomes
We need “at least one head”. Every outcome containing one head or two heads qualifies:

$$F = \{\text{HH},\;\text{HT},\;\text{TH}\}$$

Thus, the number of favourable outcomes is $$n(F)=3$$.

Step 3 – Calculate the probability
By the classical definition,

$$P(\text{at least one head}) = \dfrac{n(F)}{n(S)} = \dfrac{3}{4}$$

Therefore, the probability that Harpreet gets at least one head is $$\dfrac{3}{4}$$.

Step 1 — Describe the random experiment

The music can stop at any instant from the moment it is started until exactly 2 minutes have passed. Hence the possible stopping times (in minutes) form the continuum

$$0 \le t \le 2$$

This interval $$[0,2]$$ is the sample-space for the experiment.

Step 2 — Assumption of equal likelihood

Because the announcer has been told to stop the music "at any time" within the 2-minute period, every instant in $$[0,2]$$ is taken to be equally likely. For such a continuous and uniform situation, the probability of an event is

$$\text{Probability} \,=\, \dfrac{\text{length of favourable time-interval}}{\text{length of total time-interval}}$$

Step 3 — Identify the favourable event

We want the music to stop within the first half-minute, i.e. during

$$0 \le t \le 0.5$$

The length of this interval is

$$0.5 \text{ minutes}$$

Step 4 — Compute the probability

Total length of the sample-space interval:

$$2 \text{ minutes}$$

Therefore

$$P(\text{stop within first half-minute}) = \dfrac{0.5}{2} = \dfrac{1}{4}$$

Step 5 — Convert to decimal (optional)

$$\dfrac{1}{4} = 0.25$$

Thus the required probability is $$\dfrac{1}{4}$$ (or 25 %).

Step 1 : Identify the required areas

The helicopter is equally likely to have crashed anywhere in the entire rectangular region.
Hence the theoretical probability is

$$P(\text{crash in lake})=\frac{\text{area of lake}}{\text{area of rectangle}}.$$

Dimensions from Fig. 14.2

• Rectangle: $$200\,\text{km}\times150\,\text{km}$$
• Lake (an ellipse): major axis $$100\,\text{km}$$, minor axis $$50\,\text{km}$$

Step 2 : Calculate the area of the rectangle

$$\text{Area}_{\text{rect}} = 200\times150 = 30000\;\text{km}^2$$

Step 3 : Calculate the area of the lake (ellipse)

Semi-major axis $$a = \dfrac{100}{2}=50\,\text{km}$$
Semi-minor axis $$b = \dfrac{50}{2}=25\,\text{km}$$

Area of an ellipse: $$\pi a b$$

$$\text{Area}_{\text{lake}} = \pi\times50\times25 = 1250\pi\;\text{km}^2$$

Step 4 : Compute the probability

$$P(\text{crash in lake}) = \frac{1250\pi}{30000} = \frac{\pi}{24}$$

Step 5 : Give the numerical approximation

Using $$\pi\approx3.14$$:

$$P\approx\frac{3.14}{24}\approx0.131\;(\text{about }13.1\%)$$

Thus the probability that the helicopter crashed inside the lake is $$\frac{\pi}{24}\approx0.131$$.

Step 1 : Identify the experiment
Tossing a coin once is a random experiment. The set of all possible outcomes (sample space) is $$S = \{H,\;T\}$$, where $$H$$ stands for heads and $$T$$ stands for tails.

Step 2 : Check whether outcomes are equally likely
For a well-made, unbiased coin the physical design is symmetric, so there is no reason for either face to turn up more often in the long run. Hence each outcome is equally likely:

• Probability of heads  $$P(H)=\dfrac{1}{2}$$
• Probability of tails  $$P(T)=\dfrac{1}{2}$$

Step 3 : Relate the coin outcomes to the two teams
Before the toss, one team is asked to choose heads (or the referee assigns heads to Team A and tails to Team B). Because the outcomes are equally likely, the probability that either team wins the toss is also $$\dfrac{1}{2}$$.

Step 4 : Interpret “fairness”
A method is called fair when all competitors have the same chance of success. Since each team’s chance of getting the ball is $$\dfrac{1}{2}$$, neither team is favoured. Therefore tossing an unbiased coin is a fair way to decide which team starts the game.

Concept Used
For any event $$E$$, its probability $$P(E)$$ always satisfies the inequality
$$0 \le P(E) \le 1$$.

Check each option

• (A) $$\dfrac{2}{3}=0.666\ldots$$   lies between 0 and 1 ⇒ possible.
• (B) $$-1.5$$   is less than 0 ⇒ impossible.
• (C) $$15\% = \dfrac{15}{100}=0.15$$   lies between 0 and 1 ⇒ possible.
• (D) $$0.7$$   lies between 0 and 1 ⇒ possible.

Conclusion
The only value that does not satisfy $$0 \le P(E) \le 1$$ is $$-1.5$$.

Let the complement of event E be denoted by $$\overline{E}$$ (read “not E”).

For any event, the probabilities of the event and its complement add up to $$1$$:

$$P(E) + P(\overline{E}) = 1$$

Rearranging, we get the complement rule:

$$P(\overline{E}) = 1 - P(E)$$

The question gives $$P(E) = 0.05$$, so

$$P(\overline{E}) = 1 - 0.05 = 0.95$$

Hence, the probability of “not E” is $$0.95$$.

Step 1 – Identify the two complementary events

• Event $$E$$ : “Out of the three students, at least two have the same birthday.”
• Event $$\overline{E}$$ : “No two students have the same birthday.” (or, equivalently, “Two students do not have the same birthday”).

Since one (and only one) of these two events must occur, they are complementary, so

$$P(E) + P(\overline{E}) = 1$$

Step 2 – Insert the given probability

We are told that $$P(\overline{E}) = 0.992$$. Substituting,

$$P(E) + 0.992 = 1$$

Step 3 – Solve for $$P(E)$$

$$P(E) = 1 - 0.992 = 0.008$$

Step 4 – Write the result clearly

The probability that at least two of the three students have the same birthday is therefore

$$P(E) = 0.008 = \dfrac{8}{1000} = \dfrac{1}{125}$$.

Given data
Total number of balls in the bag = $$3+5=8$$.
Number of red balls = $$3$$.
Number of black balls = $$5$$.

Recall the definition of probability
For any event $$E$$, $$P(E)=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$$.

(i) Probability that the ball drawn is red
Favourable outcomes = number of red balls = $$3$$.
Total possible outcomes = total number of balls = $$8$$.
Therefore,
$$P(\text{red})=\dfrac{3}{8}$$.

(ii) Probability that the ball drawn is not red
“Not red” means the ball is black.
Favourable outcomes = number of black balls = $$5$$.
Total possible outcomes = $$8$$.
Hence,
$$P(\text{not red})=\dfrac{5}{8}$$.
(You could also use the complement rule: $$P(\text{not red})=1-P(\text{red})=1-\frac{3}{8}=\frac{5}{8}$$, which gives the same result.)

First write down the information given in the question.

• Number of 50 p coins  = $$100$$
• Number of ₹1 coins  = $$50$$
• Number of ₹2 coins  = $$20$$
• Number of ₹5 coins  = $$10$$

Total number of coins in the piggy bank:

$$100 + 50 + 20 + 10 = 180$$

Because the piggy bank is turned upside down and each coin is equally likely to fall out, the sample space has $$180$$ equally likely outcomes.

1. Probability that the coin will be a 50 p coin

Number of favourable outcomes (choosing a 50 p coin) = $$100$$.

Probability  = $$\dfrac{\text{favourable outcomes}}{\text{total outcomes}} = \dfrac{100}{180}$$

Simplify the fraction:

$$\dfrac{100}{180} = \dfrac{10}{18} = \dfrac{5}{9}$$

2. Probability that the coin will not be a ₹5 coin

Coins that are not ₹5 coins are the 50 p, ₹1 and ₹2 coins.

Number of such coins = $$100 + 50 + 20 = 170$$

Probability  = $$\dfrac{170}{180}$$

Again simplify:

$$\dfrac{170}{180} = \dfrac{17}{18}$$

Thus, the required probabilities are:

• (i) $$\dfrac{5}{9}$$
• (ii) $$\dfrac{17}{18}$$

The tank contains two kinds of fish.

• Male fish : $$5$$
• Female fish : $$8$$

Total number of fish in the tank

$$\text{Total} = 5 + 8 = 13$$

When one fish is taken out at random, every fish is equally likely to be chosen.

Number of favourable outcomes (choosing a male fish) is

$$\text{Favourable cases} = 5$$

By the definition of probability for equally likely outcomes,

$$P(\text{male fish}) = \dfrac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}$$

Substituting the numbers,

$$P(\text{male fish}) = \dfrac{5}{13}$$

Thus, the probability that the fish taken out is a male fish is $$\dfrac{5}{13}$$.

First identify how many pens there are altogether and how many of these are the kind we want (good pens).

Total number of pens

The collection contains $$12$$ defective pens and $$132$$ good pens.

Hence, total pens  $$= 12 + 132 = 144$$.

Favourable outcomes

We want the chosen pen to be good. The number of good pens is $$132$$.

Probability concept

When every pen is equally likely to be taken, the probability of an event is

$$\text{Probability(Event)} = \dfrac{\text{Number of favourable outcomes}}{\text{Total number of equally likely outcomes}}.$$

Substitute the relevant numbers:

$$\text{Probability(choosing a good pen)} = \dfrac{132}{144}.$$

Simplify the fraction

Both numerator and denominator are divisible by $$12$$:

$$\dfrac{132 \div 12}{144 \div 12} = \dfrac{11}{12}.$$

Therefore, the probability that the pen drawn at random is good equals $$\dfrac{11}{12} \; (\approx 0.9167).$$

Let the experiment be “drawing one disc from the box”.
Since the discs are numbered from $$1$$ to $$90$$, the sample-space contains
$$90$$ equally likely elementary events.

Total number of possible outcomes $$= 90$$.

Recall that for any event

$$P(\text{event})=\dfrac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}.$$

(i) The disc bears a two-digit number

• The smallest two-digit natural number is $$10$$.
• The greatest number on the discs is $$90$$, which is also a two-digit number.
• Thus the favourable numbers are $$10,11,12,\ldots ,90$$.

Count of those numbers:

$$\;\;90-10+1 = 81$$

Hence

$$P(\text{two-digit}) = \dfrac{81}{90} = \dfrac{9}{10}.$$

(ii) The disc bears a perfect square number

List all perfect squares not exceeding $$90$$:

$$1^2=1,\;2^2=4,\;3^2=9,\;4^2=16,\;5^2=25,\;6^2=36,\;7^2=49,\;8^2=64,\;9^2=81$$

The next square, $$10^2=100$$, is greater than $$90$$, so we stop at $$81$$.

Number of favourable outcomes $$= 9$$.

Therefore

$$P(\text{perfect square}) = \dfrac{9}{90} = \dfrac{1}{10}.$$

(iii) The disc bears a number divisible by $$5$$

• The smallest positive multiple of $$5$$ in the range is $$5$$.
• The greatest multiple of $$5$$ not exceeding $$90$$ is $$90$$ itself.

These form an arithmetic progression $$5,10,15,\ldots ,90$$ with first term $$a=5$$, common difference $$d=5$$.

If there are $$n$$ such terms, then $$a+(n-1)d = 90$$:

$$5+(n-1)\times5 = 90 \;\Rightarrow\; 5(n-1)=85 \;\Rightarrow\; n-1=17 \;\Rightarrow\; n=18.$$

So, favourable outcomes $$=18$$.

Hence

$$P(\text{divisible by }5)=\dfrac{18}{90}=\dfrac{1}{5}.$$

Therefore, the required probabilities are:

• (i) $$\dfrac{9}{10}$$
• (ii) $$\dfrac{1}{10}$$
• (iii) $$\dfrac{1}{5}$$

Step 1 – Describe the sample space
The die has six faces marked

Step 2 – Find the probability of getting A
Faces that show A: the 1st and 6th faces ⇒ 2 favourable outcomes.
Therefore
$$n(E_1)=2,$$ where $$E_1$$ is the event “getting A”.
Probability
$$P(A)=\frac{n(E_1)}{n(S)}=\frac{2}{6}=\frac13.$$

Step 3 – Find the probability of getting D
Faces that show D: only the 4th face ⇒ 1 favourable outcome.
Thus $$n(E_2)=1,$$ where $$E_2$$ is the event “getting D”.
Probability
$$P(D)=\frac{n(E_2)}{n(S)}=\frac{1}{6}.$$

Result
(i) $$P(A)=\dfrac13$$
(ii) $$P(D)=\dfrac16$$

Step 1 : Idea of geometric probability
When every point of a region is equally likely to be hit, the probability that the point will fall inside any sub-region equals the
ratio

$$\text{Probability} = \dfrac{\text{area of the favourable region}}{\text{area of the whole region}}$$

Here the whole region is the rectangle of Fig. 14.6, while the favourable region is the circle drawn inside it.

Step 2 : Area of the rectangular region
From the figure, the rectangle measures $$3\,\text{m}$$ by $$2\,\text{m}$$.

$$A_{\text{rect}} = l \times b = 3\,\text{m} \times 2\,\text{m} = 6\,\text{m}^2$$

Step 3 : Area of the circular region
The circle has diameter $$1\,\text{m}$$, so its radius is

$$r = \dfrac{1\,\text{m}}{2} = 0.5\,\text{m}$$

Hence

$$A_{\text{circle}} = \pi r^2 = \pi(0.5)^2 = 0.25\,\pi\;\text{m}^2$$

Step 4 : Required probability

$$P = \dfrac{A_{\text{circle}}}{A_{\text{rect}}} = \dfrac{0.25\,\pi}{6} = \dfrac{\pi}{24}$$

Taking $$\pi \approx 3.14$$,

$$P \approx \dfrac{3.14}{24} \approx 0.13$$

Therefore the probability that the die will land inside the circle is $$\dfrac{\pi}{24}\; (\text{about }13\%).$$

Given: A fair one–rupee coin is tossed $$3$$ times. All $$2^3 = 8$$ possible sequences of Heads (H) and Tails (T) are equally likely.

Hanif wins only when the three results are identical:

• All Heads  $$\to$$  H H H
• All Tails  $$\to$$  T T T

So, number of winning outcomes $$= 2.$$

Therefore, number of losing outcomes

$$8 - 2 = 6.$$

The required probability is

$$P(\text{Hanif loses}) = \dfrac{\text{number of losing outcomes}}{\text{total outcomes}} = \dfrac{6}{8} = \dfrac{3}{4}.$$

Experiment
Throwing a die twice (or, equivalently, throwing two dice together) gives a sample space of size
$$6 \times 6 = 36$$ equally likely ordered pairs.

Favourable outcomes

• A single throw produces the face $$5$$ with probability $$\dfrac16$$.
• It produces a face other than $$5$$ with probability $$\dfrac56$$.

(i) “5 will not come up either time”

The two throws are independent, so

$$P(\text{no 5 in first throw}) = \frac56$$
$$P(\text{no 5 in second throw}) = \frac56$$

Hence

$$P(\text{no 5 in both throws}) = \frac56 \times \frac56 = \frac{25}{36}$$

(ii) “5 will come up at least once”

“At least once” is the complement of “never”. Therefore

$$P(\text{at least one 5}) = 1 - P(\text{no 5 in both throws})$$
$$= 1 - \frac{25}{36} = \frac{11}{36}$$

Verification by direct counting (optional)

The three mutually exclusive cases are:

• $$5$$ on the first throw, not $$5$$ on the second: $$\frac16 \times \frac56 = \frac5{36}$$
• Not $$5$$ on the first, $$5$$ on the second: $$\frac56 \times \frac16 = \frac5{36}$$
• $$5$$ on both throws: $$\frac16 \times \frac16 = \frac1{36}$$

Total $$= \frac5{36}+\frac5{36}+\frac1{36}=\frac{11}{36}$$, confirming the previous result.