Advance Math Model Questions Set 2 Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (b)

For each person to know all the secrets the communication has to be between the Englishmen (who knows say E1 French) and one Frenchmen (say F1). The other two in each case will communicate with E1 & F1 respectively. So for minimum no. of calls, E2 gives information to E1 & receives it after E1 interacts with F1. So 2 calls for each of the four E2, E3, F2 and F3, i.e., 8 calls +1 call (between E1 & F1). Hence 9 calls in all.

Answer: (c)

Total number of attempts = 20

Favourable no. of attempts = 5

Required probability (running the program correctly in the third run) = $5/{20} = 1/4$

Answer: (a)

Take, A and B to be always together as a single entity.

Now, total no. of children = 4 – 2 + 1 = 3

These can be arranged in 3! ways and A, B can be arranged among themselves in 2! ways.

Hence, no. of arrangements such that A and B are always together = 3! × 2! = 3 × 2 × 2 = 12

Answer: (d)

Probability that first ball is white and second black

= (4/6) × (5/8) = 5/12

Probability that first ball is black and second white

= (2/6) × (3/8) = 1/8

These are mutually exclusive events hence the required probability

P = $5/{12} + 1/8 = {13}/{24}$.

Answer: (c)

As there are 6 letters and envelopes, so if exactly 5 are into correctly addressed envelopes, then the remaining 1 will automatically be placed in the correctly addressed envelope. Thus, the probability that exactly 5 go into the correctly addressed envelope is zero.

Answer: (c)

Since, A and B are independent events.

∴ P(A ∩ B) = P(A)P(B)

Further since, A ∩ C, B ∩ C, A ∩ B ∩ C are subsets of C, we have

P(A ∩ C) ≤ P(C) = 0

P(B ∩ C) ≤ P(C) = 0

and P(A ∩ B ∩ C) ≤ P(C) = 0

⇒ P(A ∩ C) = 0 = P(A)P(C)

P(B ∩ C) = 0 = P(B)P(C)

P(A ∩ B ∩ C) = 0 = P(A)P(B)P(C).

Clearly A, B, C are pairwise independent as well as mutually independent. Thus, A,B,C are independent events.