Quant Quiz for SBI PO 2017 (Probability and P&C)

Quant Quiz for SBI PO 2017 (Probability and P&C)

1. A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?

a) 1260

b) 210

c) 210 × 6!

d) 1512

e) None of these

Answer: A)

A team of 6 members has to be selected from the 10 players. This can be done in 10C6 or 210 ways.

Now, the captain can be selected from these 6 players in 6 ways.

Therefore, total ways the selection can be made is 210×6= 1260

2. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

a) 564

b) 735

c) 756

d) 657

e) 854

Answer: C)

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

Required number of ways= (⁷C₃ × ⁶C₂) + (⁷C₄ × ⁶C₁) + ⁷C₅ = 756.

3. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

a) 25200

b) 52000

c) 120

d) 24400

e) 12000

Answer: A)

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (⁷C₃ × ⁴C₂) = 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves = 5! = 120

Required number of ways = (210 x 120) = 25200.

4. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

a) 209

b) 290

c) 200

d) 208

e) 280

Answer: A)

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

Required number of ways = (⁶C₁ × ⁴C₃) + (⁶C₂ × ⁴C₂) + (⁶C₃ × ⁴C₁) + ⁶C₄ = 209.

5. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

a) 48

b) 64

c) 63

d) 45

e) 52

Answer: B)

We may have (1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

Required number of ways = ³C₁ × ⁶C₂ + (³C₂ × ⁶C₁) + ³C₃ = (45 + 18 + 1) = 64

6. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

a) ½

b) 3/5

c) 9/20

d) 8/15

e) 7/18

Answer: C)

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/n(S) = 9/20.

7. In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

a) 1/3

b) 3/5

c) 8/21

d) 7/21

e) 9/21

Answer: A)

Total number of balls = (8 + 7 + 6) = 21.

Let E = event that the ball drawn is neither red nor green = event that the ball drawn is blue.

n(E) = 7

P(E) = n(E)/n(S) = 7/21 = 1/3

8. A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

a) 10/21

b) 11/21

c) ½

d) 2/7

e) 3/7

Answer: A)

Total number of balls = (2 + 3 + 2) = 7

Let S be the sample space.

Then, n(S) = Number of ways of drawing 2 balls out of 7 =7C2 = 21

Let E = Event of drawing 2 balls, none of which is blue.

n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls = 5C2 = 10

Therefore, P(E) = n(E)/n(S) = 10/ 21

9. In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

a) 21/46

b) 1/5

c) 3/25

d) 1/50

e) 2/25

Answer: A)

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting 3 students out of 25 = 25C3 = 2300.

n(E)= 10C1 × 15C2 = 1050. 

P(E) = n(E)/n(s) = 1050/2300 = 21/46

10. A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that the problem will be solved?

a) ¼

b) ½

c) ¾

d) 7/12

e) 5/12

Answer: C)