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

Answer: (a)

P(x) = 0.7

= The probability of selected in firm X

P(y) = 1 – 0.5 = 0.5

= probability of selected in firms Y

P(X ∪ Y) = 1 – 0.6 = 0.4

Probability that he selected in one of the firms

= 0.7 + 0.5 – 0.4 = 0.8

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.

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: (b)

Suppose three people have been given a, b and c number of items.

Then, a × b × c = 30

Now, There can be 5 cases :

Case I : When one of them is given 30 items and rest two 1 item each.

So, number of ways for (30 × 1 × 1) = ${3!}/{2!}$ = 3

(As two of them have same number of items)

Case II : Similarly, number of ways for (10 × 3 × 1) = 3! = 6

Case III : Number of ways for (15 × 2 × 1) = 3! = 6

Case IV : Number of ways for (6 × 5 × 1) = 3! = 6

Case V : Number of ways for (5 × 3 × 2) = 3! = 6

Here, either of these 5 cases are possible.

Hence, total number of ways = 3 + 6 + 6 + 6 + 6 = 27

Answer: (c)

The total possible pairs of children (B, B), (B, G), (G, B). Now the one child is boy, is confirmed, but we don't know whether he is youngest or elder one. So the three ordered pairs could be the one describing the children in this family. So the probability of the younger children to be boy = 2/3.

Answer: (d)

To construct 2 roads, three towns can be selected out of 4 in 4 ×3×2 = 24 ways. Now if the third road goes from the third town to the first town, a triangle is formed, and if it goes to the fourth town, a triangle is not formed. So there are 24 ways to form a triangle and 24 ways of avoiding a triangle.