MAH MBA CET Probability and Combinatorics Questions

When tossing 3 coins, each coin has 2 possibilities (H or T). The total number of possible outcomes is 2^3 = 8.

The sample space (S) is:

HHH
HHT
HTH
THH
HTT
THT
TTH
TTT

The condition is "at least two" tails, which means we are looking for outcomes with exactly 2 tails or exactly 3 tails. 

Exactly 2 tails: {HTT, THT, TTH} — 3 outcomes 

Exactly 3 tails: {TTT} — 1 outcome

Total favorable outcomes = 3 + 1 = 4

Hence:

P(Atleast 2 tails) = 4/8 = 1/2 = 50%

The probability that the first ball is not yellow is $$\frac{7}{12}$$
The probability that the second ball is not yellow is $$\frac{6}{11}$$
Hence, the total probability is $$\frac{7}{12}*\frac{6}{11}=\frac{7}{22}$$

Number of favourable cases = 6
Total number of cases = 6*6 = 36
So, probability = 1/6

Total number of letters = 6 of which 3 are the same.

Therefore, number of ways of arranging = $$\frac{6!}{3!}$$ = $$\frac{720}{6}$$ = 120

MIRACLE contains I,A,E as vowels apart from M, R, C, L as consonants. Group I, A, E as one unit.
Total number of ways = 5!
Now I, A, E can be arranged in 3! ways.
Total number of ways = 5!*3! = 120*6 = 720 ways

Assign Task 2 first: 2 choices (Person 3 or 4).

Total ways to assign the remaining 5 tasks: 5! = 120.

Total without Task 1 restrictions: 2*120 = 240.

Subtract restricted cases for Task 1:

If Task 1 is given to Person 1: 2*1*4! = 48.

If Task 1 is given to Person 2: 2*1*4! = 48.

Final Result: 240 - 48 - 48 = 144.

We want to minimise the total number of coins under these conditions:

7 pockets, each has at least 1 coin

At most 3 pockets can have the same number of coins

The remaining pockets must all have distinct numbers

To minimise total coins: we can use the smallest possible positive integers and allow exactly 3 pockets to have the same number (since “at most 3” → we use 3 to minimise total). The other 4 pockets must all be distinct and different from the repeated number.

Let the repeated number be 1 (smallest possible): so 3 pockets: 1, 1, 1

Now choose 4 distinct numbers, all different from 1: Next smallest distinct numbers: 2, 3, 4, 5

Hence, Total coins: 1 + 1 + 1 + 2 + 3 + 4 + 5 = 17

Let 

P(Alka selected) = 1/7 so P(Alka not selected) = 6/7

P(Alkit selected) = 1/5 sp P (Alkit not selected) = 4/5

So, the probability of only one being selected: 

Case 1: Alka selected, Alkit not selected

$$\dfrac{1}{7}\times\dfrac{4}{5}=\dfrac{4}{35}$$

Case 2: Alka not selected, Alkit selected

$$\dfrac{6}{7}\times\dfrac{1}{5}=\dfrac{6}{35}$$

Total probability: $$\dfrac{4}{35}+\dfrac{6}{35}=\dfrac{10}{35}=\dfrac{2}{7}$$

Total possible combinations when 2 dice are thrown=6*6=36
Product of 2 numbers is even when either both of them are even or one is even, other is odd.
There are 3 even no.(2,4,6) and 3 odd no.(1,3,5) possible in a dice.
Total combinations when both are even = 3*3 = 9
Total combinations when one is even and other is odd = 3*3(first no. is odd, second even) + 3*3(first no. is even, second odd) = 18
Required probability = $$\frac{\ 9+18}{36}=\ \frac{\ 27}{36}=\ \frac{\ 3}{4}\ $$
Option E is the answer.
 

Let's say Prakhar has only one coin with heads on both sides. If there is only one such coin, then there would be cases in which he chooses 4 out of those six coins, but he does not pick that coin 
Then, his probability of getting at least one head can not be 1. 

For this probability to be 1, he must have at least three coins with heads on both sides. So, no matter what four coins he picks, he picks one coin with heads on both sides, resulting in his probability of getting at least one head 1. 

The same logic applies to tails; he must have three coils with tails on both sides. 

Therefore, all six coins with prakhar have either heads or tails on both sides. 

Hence, Option E is the correct answer. 

The question uses the Bayes theorem., but it can also be solved by thinking logically 

Essentially, we are given that it rains, and when it rains, Ramya brings the umbrella with a probability of 0.6. Hence the probability that she does not bring the umbrella on Monday is 0.4

These are two independent events and hence do not affect each other. 
The question states that is rains; the probability of Ramy not having an umbrella is independent of this and can directly be answered to be 0.4

Using Bayes theorem too, we get the same answer:$$\frac{0.4\times\ 0.6}{0.4\times\ 0.6\ +\ 0.6\times\ 0.6}=0.4$$

Hence, Option A is the correct answer. 

The number of ways of distributing $$n$$ items among $$r$$ persons is $$^{n+r-1}C_{r-1}$$

If at least one item is given to each person, then the number of ways is $$^{n-1}C_{r-1}$$

As there are five chocolates and three children, the number of ways is $$^{n+r-1}C_{r-1}=^{5+3-1}C_{3-1}=^7C_2=21$$

For cookies, there are six cookies and we need to give at least 1 to each child, the number of ways is $$^{n-1}C_{r-1}=^{6-1}C_{3-1}=^5C_2=10$$

So, the total number of ways is $$21\times10=210$$

Hence, the answer is 210.

There are eight corners in the die; we can choose any of them in 8 ways. 

Say we choose the corner with numbers (3,5,1); now we have to paint them in colours red, blue and green. 
This again can be done in 3! ways. 

The remaining three faces are to be painted with the remaining three colours, which again can be done in three ways! 

Giving us a total of $$8\times\ 6\times\ 6=288$$

Therefore, Option E is the correct answer.