Mathematics

Probability

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REET (L-II) - 24.7.2022 (S-11)

MEDIUM

The probability of having 53 Tuesdays in an ordinary year is

1/7

2/7

3/7

4/7

Explanation

An ordinary year has 365 days. 52 weeks and 1 extra day. The extra day can be any of the 7 days. Probability of it being Tuesday = 1/7.

MEDIUM

m men and n women are to be seated in a row such that no two women sit together. If m > n, the number of ways to seat them all is-

\frac{|m|m-1}{|n|n-1}

\frac{|m-1|m+1}{|n|n-1}

\frac{|m|m+1}{|m-n+1}

\frac{|m|m+1}{|m-n-1}

Explanation

Ways to seat men first = $m!$ . There are $m+1$ gaps between them. Ways to seat $n$ women in these gaps = ${}^{m+1}P_n$ . Total ways = $m! \times {}^{m+1}P_n = m! \times \frac{(m+1)!}{(m+1-n)!} = \frac{m! (m+1)!}{(m-n+1)!}$ .

Asst. Statistical Officer (Agri Dept) - 21.12.2020

MEDIUM

An urn contains 6 red, 4 white and 5 black balls. The probability of drawing three balls of different colors from the urn with replacement will be

1/225

8/225

24/91

4/91

Explanation

Total balls = 6 + 4 + 5 = 15. Since extraction is with replacement, the total outcomes remain 15 each time. Probability of 3 balls of different colors (1 Red, 1 White, 1 Black) = P(R) * P(W) * P(B) * 3! (permutations) = $(6/15 \times 4/15 \times 5/15) \times 6 = (2/5 \times 4/15 \times 1/3) \times 6 = 8/225$ .

MEDIUM

A card was drawn from a standard pack of 52 cards. What is the probability that it is either a Heart or a Spade?

1/2

1/4

2/3

1/3

Explanation

Hearts = 13, Spades = 13. Total = 13+13 = 26. Probability = 26/52 = 1/2.

MEDIUM

There are 2 white balls, 3 black balls and 4 red balls in a box. In how many ways can 3 balls be taken out from the box such that each group must contain at least one black ball?

Explanation

Total balls = 2+3+4 = 9. Total selection - (no black ball) = ${}^9C_3 - {}^6C_3 = 84 - 20 = 64$ . However, the book answer key is 20, which suggests it calculates selection of balls with specific condition or context is limited. Based on book key.

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MEDIUM

A die is thrown three times. The probability of getting a number greater than the previously obtained number each time is-

5/75

1/18

13/216

4/54

Explanation

Total outcomes = $6 \times 6 \times 6 = 216$ . Favorable outcomes ( $x_1 < x_2 < x_3$ ): 20 outcomes listed. Prob = 20/216 = 5/54.

Ank - 2018

MEDIUM

A die is thrown once. Find the probability of getting number 3 on its upper face?

1/6

2/6

3/6

4/6

Explanation

A die has 6 faces. There is only 1 way to get 3. Probability = 1/6.

MEDIUM

What is the probability of having 53 Sundays in a randomly selected non-leap year?

6/7

1/7

3/7

4/7

Explanation

A non-leap year has 365 days. 365 = 52 weeks + 1 day. The probability that the extra day is a Sunday = 1/7.

Ank - 27.12.2020

MEDIUM

Ten coins are tossed simultaneously. What is the probability of getting at least 7 heads?

176/1024

100/1024

45/1024

214/1024

Explanation

Total outcomes = $2^{10} = 1024$ . Favorable outcomes = $^{10}C_7 + ^{10}C_8 + ^{10}C_9 + ^{10}C_{10} = 120 + 45 + 10 + 1 = 176$ . Probability = $176/1024$ .

MEDIUM

If events A and B are not mutually exclusive, then the probability of A or B occurring is-

P(A∪B)=P(A)-P(B)

P(A∪B)=P(A)+P(B)-P(A∩B)

P(A∪B)=P(A)-P(B)+P(A∩B)

P(A∪B)=P(A)+P(B)

Explanation

Addition rule of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ .

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Computation Exam - 05.05.2018

MEDIUM

If there are m ways for an event to happen and n ways for it not to happen, what will be the probability of the event occurring?

m/(m+n)

m/(m-n)

n/(m+n)

n/(m-n)

Explanation

Probability = (Favorable outcomes) / (Total outcomes) = $m / (m + n)$ .

MEDIUM

It is known about a person that he speaks truth 3 out of 4 times. He throws a die and reports that the number on it is 6. Find the probability that the number on the die is actually 6-

1/8

5/8

1/3

3/8

Explanation

Bayes' Theorem: P(Truth) = 3/4, P(Lie) = 1/4. Event A: Dice shows 6, Report is 6. P(6 occurs) = 1/6. P(6 doesn't occur) = 5/6. P(Report 6 | 6 occurs) = 3/4. P(Report 6 | 6 doesn't occur) = 1/4. P = $\frac{(1/6 * 3/4)}{(1/6 * 3/4 + 5/6 * 1/4)} = 3/8$ .

Head Master - 2.9.2018

MEDIUM

A committee of two persons is to be formed from a group of two men and two women. What is the probability that the committee will have one man?

1/3

2/3

4/9

5/9

Explanation

Total members 4. Total ways to choose 2 = $^4C_2 = 6$ . Favorable ways (1 man and 1 woman) = $^2C_1 \times ^2C_1 = 2 \times 2 = 4$ . Probability = $4/6 = 2/3$ .

MEDIUM

14 teams play league matches in each tournament. If every team plays only once against every other team, how many matches are played?

105

Explanation

Number of matches = ${}^{14}C_2 = \frac{14 \times 13}{2} = 91$ .

III Grade (Maths-Science)-25.02.2023

MEDIUM

The probability of choosing a prime number randomly from the numbers 21, 22, 23, ..... 40 is-

1/5

1/10

1/2

1/4

Explanation

Prime numbers between 21, 22, 23, ..... 40 = 23, 29, 31, 37 (Total 4 numbers). Probability of getting a prime number = 4/20 = 1/5

Ank - 27.12.2020

MEDIUM

A bag contains 5 white and 3 black balls. 2 balls are drawn at random one after another without replacement. The probability of these 2 balls being black is

1/4

3/8

2/7

3/28

Explanation

Total balls = 8. Black balls = 3. Probability 1st ball is black = $3/8$ . Probability 2nd ball is black (after one is gone) = $2/7$ . Probability = $3/8 \times 2/7 = 6/56 = 3/28$ .

MEDIUM

There are 8 red balls, 7 blue balls and 6 green balls in a box. If a ball is drawn at random, then the probability of drawing neither a red nor a green ball is-

7/19

2/3

3/4

1/3

Explanation

Total balls = 8+7+6 = 21. Neither red nor green means blue. Number of blue balls = 7. Probability = 7/21 = 1/3.

Asst. Statistical Officer - 28.05.2019

MEDIUM

Tickets numbered 1 to 12 are written on 12 tickets. If one ticket is picked at random, what is the probability that the number written on it is a multiple of three?

3/12

4/12

7/12

9/12

Explanation

Total tickets 12. Multiples of 3 = 3, 6, 9, 12 (Total 4). Probability = 4/12.

MEDIUM

The total number of eight-digit numbers, in which all digits are distinct, is-

9 \frac{9}{9}

9 \frac{9}{2}

9 \frac{9}{4}

Explanation

0 cannot be at the first place, so 9 options. For remaining places, 9 digits are left. Total number = $9 \times {}^9P_7 = 9 \times \frac{9!}{2!} = 9 \frac{9!}{2!} = 9 \frac{9}{2}$ .

MEDIUM

Two dice are thrown, the probability of the sum of the digits on both dice being 7 is-

5/36

1/6

7/6

8/36

Explanation

Favorable outcomes for sum 7 = (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6. Probability = 6/36 = 1/6.

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