Probability Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (d)

Total no. of balls =4 + 3 + 5 = 12

n(S) = $^{12}C_2$ = ${12 × 11}/2$ = 66

n(E) = $ ^4C_2+ ^3C_2 + ^5C_2 $ = ${{4 × 3}/2} + {{3 × 2}/2} + {{5 × 4}/2}$

= 6 + 3 + 10 = 19

∴ Required probability, P(E) = ${n(E)}/{n(S)}$ = ${19}/{66}$

Answer: (d)

For 1st student, Probability of selecting any one day as his birthday = ${365}/{365}$ = 1 Now, the remaining two students to be selected must have same day as their birthday as for the 1st student.

Probability of rest two students, having the same birthday as

that of the 1st student = $1/{365}$ × $1/{365}$

Hence, required probability = 1 × $1/{(365)}^2$ = $1/{(365)^2}$

Answer: (b)

Total number of caps = 12

n(S) = $^{12}C_3$ = ${12!}/{3! × 12! -3}$ = ${12 × 11 × 10 × 9!}/{3 × 2 × 1 × 9!}$ = 220

n$(E_1)$ = Out of 4 red caps, number of ways to pick 2 caps = $^4C_2$

= ${4!}/{2! × 4! -2}$ = ${4 × 3 × 2 × 1}/{2 × 1 × 2 × 1}$ = 6.

n$(E_2)$ = Out of 5 green caps, number of ways to pick one

cap = $^5C_1$ = 5

p(E) = ${n(E_1) × n(E_2)}/{n(S)}$ = ${6 × 5}/{220}$ = ${3}/{22}$

Answer: (d)

Probability of a particle lying in any particular half = $1/2$

∴ Probability of all 10 particles lying in either 1st half or

2nd half = $(1/2)^{10} + (1/2)^{10}$ = 2$(1/2)^{10}$ = $2^{1/9}$

Answer: (d)

Selection of 1 boy and 3 girls in $^5 C_1$ × $^4 C _3$ = 5 × 4 = 20 ways

Selection of 4 girls and no boy in $^5 C_0$ × $^4 C _4$ = 1 × 1 = 1 way

∴ n(E) = total no. of ways = 21

Without any restriction, a committee of 4 can be formed from among 4 girls and 5 boys in $^9C_4$

= ${9 × 8 × 7 × 6}/{4 × 3 × 2}$ = 9 × 7 × 2 ways

∴ P(E) = ${n(E)}/{n(S)}$ = ${21}/{9 × 7 × 2}$ = $1/6$