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

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.

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: (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: (a)

Suppose $E_1 , E_2$ and $E_3$ are the events of winning the race by the horses A, B and C respectively

∴ $P(E_1) = 1/{1 + 3} = 1/4, P(E_2) = 1/{1 + 4} = 1/5$

$P(E_3) = 1/{1 + 5} = 1/6$

∴ Probability of winning the race by one of the horses A, B and C

= $P (E_1 or E_2 or E_3) = P(E_1) + P(E_2) + P(E_3)$

= $1/4 + 1/5 + 1/6 = {37}/{60}$

Answer: (b)

Probability of getting a diamond, P(D) = ${13}/{52} = 1/4$

and probability to king, P(K) = $4/{52} = 1/{13}$

So, required probability = P(D).P(K)

= $1/4 × 1/{13} = 1/{52}$

Answer: (a)

$p_1 = {^{15}C_2}/{^{42}C_4} = {15 × 14 × 4!}/{2! × 42 × 41 × 40 × 39} = 1/{41 × 26}$ and

$p_2 = {^{30}C_4}/{^{84}C_8} = {15 × 14 × 13 × 12 × 8!}/{4! × 84 × 83 × 82 × ...... × 77}$

= ${15 × 14 × 13 × 12 × 8 × 7 × 6 × 5}/{84 × 83 × 82 × 81 × 79 × 78 × 77} < p_1$

⇒$p_1 > p_2$.