Practice Questions for Modern Math - Probability

1. PE 1 - Probability | Modern Math - Probability

If 3 consecutive letters are selected at random from the English alphabet, then the probability that one of the letters is a vowel is:

(a)
7/12
(b)
5/12
(c)
5/8
(d)
None of these

2. PE 1 - Probability | Modern Math - Probability

Box A contains 4 red and 6 green balls. Box B contains 7 red and 3 green balls. A ball is drawn from box A and without seeing its colour, it is put into box B. If a ball is now drawn from box B, then the probability that it is red is:

(a)
6/11
(b)
43/55
(c)
37/55
(d)
49/5

3. PE 1 - Probability | Modern Math - Probability

The letters of the word ‘JUMBLE’ are arranged at random. Find the probability of having exactly two letters in between N and R.

(a)
1/5
(b)
1/4
(c)
1/10
(d)
1/2

4. PE 1 - Probability | Modern Math - Probability

400 tickets are numbered as 000, 001, 002, 003,……., 599. If a ticket is drawn at random from them and M is the event that the sum of the digits of the number is 6, then P(M) =

(a)
1/20
(b)
9/200
(c)
1/24
(d)
None of these

5. PE 1 - Probability | Modern Math - Probability

An unbiased dice is rolled 4 times. Out of 4 numbers that are shown up, the probability that the least is greater than 1 and the greatest is less than 5 is:

(a)
1/2
(b)
1/32
(c)
1/8
(d)
1/16

6. PE 1 - Probability | Modern Math - Probability

Saransh, Akhil and Ronak start a game of dice. They throw the dice by turns, first Saransh, Akhil then Ronak, then once again Saransh, Akhil and Ronak and so on until one of them wins. Saransh is considered to have won if he throws 1 or 2 or 3, Akhil if he throws a prime number and Ronak if he throws 4 or 5 or 6. What is the probability that Saransh wins?

(a)
3/7
(b)
4/7
(c)
1/3
(d)
Cannot be determined

Answer: Option B

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Explanation :

P(S) = P(Saransh getting 1 or 2 or 3) = $\frac{3}{6}$ = $\frac{1}{2}$
P(A) = P(Akhil getting 2 or 3 or 5) = $\frac{3}{6}$ = $\frac{1}{2}$
P(R) = P(Ronak getting 4 or 5 or 6) = $\frac{3}{6}$ = $\frac{1}{2}$

P(Saransh wins) =
P(Saransh gets 1 or 2 or 3 in his first turn) +
P(Saransh gets 1 or 2 or 3 in his second turn) +
P(Saransh gets 1 or 2 or 3 in his third turn) + …

For Saransh to win in his second turn, Akhil and Ronak must not win in their first turn. Same applies for Saransh winning in his further turns.

⇒ P(Saransh wins) = P(S) + P(S’) × P(A’) × P(R’) × P(S) + P(S’) × P(A’) × P(R’) × P(S’) × P(A’) × P(R’) × P(S) + …

= $\frac{1}{2}$ + $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$ + $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$ +⋯

= $\frac{1}{2}$ + ${(\frac{1}{2})}^{4}$ + ${(\frac{1}{2})}^{7}$ + ⋯ [Sum of an infinite GP]

= $\frac{\frac{1}{2}}{1 - \frac{1}{8}}$ = $\frac{4}{7}$ .

Hence, option (b).

7. PE 1 - Probability | Modern Math - Probability

A letter lock has 4 rings each containing six distinct letters. What is the probability that the code is HIJK in same order?

(a)
${(\frac{1}{C_{6}^{26}})}^{4}$
(b)
1/104
(c)
${(\frac{1}{26})}^{4}$
(d)
${(\frac{1}{C_{6}^{26}})}^{4} \times {(\frac{1}{6})}^{4}$

Answer: Option C

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Explanation :

Probability that H is the 1st letter of the code = $\frac{1}{26}$
Probability that I is the 2nd letter of the code = $\frac{1}{26}$
Probability that J is the 3rd letter of the code = $\frac{1}{26}$
Probability that K is the 4th letter of the code = $\frac{1}{26}$

⇒ Required probability = $\frac{1}{26}$ × $\frac{1}{26}$ × $\frac{1}{26}$ × $\frac{1}{26}$ = ${(\frac{1}{26})}^{4}$

Hence, option (c).

8. PE 1 - Probability | Modern Math - Probability

A man has five sticks each of length 1 m, 2 m, 3 m, 4 m and 5 m. He chooses three of them randomly and tries to make a triangle. What is the probability that he succeeds?

(a)
0.2
(b)
0.7
(c)
0.3
(d)
0.1

9. PE 1 - Probability | Modern Math - Probability

In a three letter password, what is the probability that all three letters are same?

(a)
6/26³
(b)
1/26²
(c)
1/36
(d)
None of these

PE 1 - Probability | Modern Math - Probability

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