Advance Math Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (a)

P (None is defective) = ${^{16}C_2}/{^{20}C_2} = ({16 × 15}/{2 × 1} × {2 × 1}/{20 × 19})$

= ${12}/{19}.$

P (at least one is defective) = 1 - ${12}/{19} = 7/{19}$

Answer: (b)

Let E be the event of selecting the three numbers such that their product is odd and S be the sample space. For the product to be odd, 3 numbers choosen must be odd.

∴ n(E) = $^5C_3$

= n(S) = $^9C_3$

∴ P(E) = ${n(E)}/{n(S)} = {^5C_3}/{^9C_3} = 5/{42}$

Answer: (c)

Let there be n participants in the beginning. Then the number of games played by (n – 2) players = $^{n - 2}C_2$

∴ $^{n - 2} C_2 + 6 = 84$

(Two players played three games each)

⇒$^{n - 2}C_2$ = 78⇒(n - 2)(n - 3) = 156

⇒$n^2$ - 5n - 150 = 0

⇒$n^2$ - 15n + 10n = 150 = 0

⇒n(n - 15) + 10(n - 15) = 0

(n - 15)(n + 10) = 0

n = 15, –10

n cannot be –ve

Therefore n = 15.

Answer: (b)

Exhaustive number of cases = $^{24}C_{14}$

Favourable cases = $^{22}C_{14}$

Probability = ${^{22}C_{14}}/{^{24}C_14} = {15}/{92}$

Answer: (b)

Required probability = $(1 - 1/6) × (1- 2/5) = 5/6 × 3/5 = 1/2$

Answer: (a)

Exhaustive no. of cases = $6^3$

10 can appear on three dice either as distinct number as following (1, 3, 6) ; (1, 4, 5); (2, 3, 5) and each can occur in 3! ways. Or 10 can appear on three dice as repeated digits as following (2, 2, 6), (2, 4, 4), (3, 3, 4) and each can occur in ${3!}/{2!}$ ways.

∴ No. of favourable cases = 3 × 3! + 3 × ${3!}/{2!}$ = 27

Hence, the required probability = ${27}/{6^3} = 1/8$