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

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: (c)

Considering the two vowels E and A as one letter, the total no. of letters in the word 'EXTRA' is 4 which can be arranged in $^4P_4$ , i.e. 4! ways and the two vowels can be arranged among themselves in 2! ways.

∴ reqd. no. = 4! × 2! = 4 × 3 × 2 × 1 × 2 × 1 = 48

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: (a)

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

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

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

⇒$p_1 > p_2$.

Answer: (a)

It can be clearly established that the choices (a) , (c) and (d) may or may not be true . Statement (b) can never be true because every person cannot have a different number of acquaintances.

Answer: (a)

Six students $S_1 , S_2 , ........, S_6$ can be arranged round a circular table in 5 ! ways. Among these 6 students there are six vacant places, shown by dots (•) in which six teachers can sit in 6 ! ways.

Hence, number of arrangement = 5 ! × 6 !