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

Answer: (a)

The 7 different letters of the word ARTICLE can be arranged in 7! ways, i.e., n(S) = 7!

n(E) = $^3 P_3$ × $^4P_4$ = 3! × 4! = 6 × 34

∴ P(E) = ${6 × 24}/{7!}$ = $1/{35}$

Answer: (c)

Total number of caps = 12

Total result n(S) = ${12} C_4$.

n(S) = ${12!}/{4! × 12! - 4}$ = ${12 × 11 × 10 × 9 × 8!}/{4 × 3 × 2 × 1 × 8!}$ = 5 × 99

n$(E_1)$ = Out of 5 caps, number of ways to not pick a green cap = $^5C_0$.

n$(E_2)$ = Out of 7 caps, number of ways to pick 4 caps

= $^7C_4$. = ${7!}/{4! × 7! - 4}$ = ${7 × 6 × 5 × 4 × 3!}/{4 × 3 × 2 × 1 × 3!}$ = 35

p(E) = ${n(E_1) n (E_2)}/{n(s)}$ = ${1 × 35}/{5 × 99}$ = ${7}/{99}$

Answer: (d)

Probability of Head or Tail on the upper side for a coin

= $1/2$ ∴ Probability of same side on the upper side for the three coins = ${1/2} × {1/2} × {1/2}$ = $(1/2)^3$ = $1/8$

∴ angles are 80°, 60° and 40°

Answer: (b)

P(E) = Probability of passing in English = 0.6

P(E ∩ M) = Probability of passing in Maths and English

= 0.54

P(M) = Probability of passing in Maths

Since, P(M) and P(E), both are independent events.

So, P(E ∩ M) = P(E) × P(M)

P(M) = P(E ∩ M)/ P(E) = ${0.54}/{0.6}$ = 0.9

∴ Probability of failing in Maths = 1 – 0.9 = 0.1 = 10%

Answer: (d)

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