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

Answer: (b)

Total possible result = Selection of 3 marbles out of 12

= $^{12}C_3$ = ${12 × 11 × 10}/{1 × 2 × 3}$ = 220

Total number of event n(E) = $^3C_3 + ^4C_3$.

n (E) = 1 + 4 = 5

Required probability = ${n(E)}/{n(S)}$ = ${5}/{220}$ = ${1}/{44}$

Answer: (e)

Total number of caps = 2 + 4 + 5 + 1 = 12

Total result n(S) = $^{12} C_ 2$.

n(S) = ${12!}/{2! × 12! - 2}$ = ${12!}/{2! × 10!}$ = ${12 × 11 × 10!}/{2 × 1 × 10!}$ = 66

Favorable result n(E) = $^2 C _2$ = 1

Required probability p(E) = ${n(E)}/{n(S)}$ = $1/{66}$

Answer: (a)

Total number of caps = 12

n (S) = $^{12}C_1$ = 12

Out of (2 blue + 1 yellow) caps number of ways to pick one cap n(E) = $^3C_1$ = 3

Required probability p(E) = ${n(E)}/{n(S)}$ = ${3}/{12}$ = $1/4$

Answer: (a)

Total number of balls = 5 + 4 + 3 = 12

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

i.e., 3 marbles out of 12 marbles can be drawn in 220 ways.

If all the three marbles are of the same colour, it can be done in $^5 C _3 + ^4 C_3 + ^3 C_3$ = 10 + 4 + 1 = 15 ways

Now, P(all the 3 marbles of the same colour) + P(all the 3 marbles are not of the same colour) = 1

∴ P(all the 3 marbles are not of the same colour)

= 1- ${15}/{220}$ = ${205}/{220}$ = ${41}/{44}$

Answer: (a)

n(S) = Number of ways to select 3 marbles out of 7 marbles = $^7C_3$.

= ${7 × 6 × 5}/{1 × 2 × 3}$ = 35

n(E) = Probability that two are green and one is red = $^4C_2$ × $^3C_1$

= ${4 × 3}/{1 × 2}$ × 3 = 18 Required probability = ${n(E)}/{n(S)}$ = ${18}/{35}$