Probability Questions for SBI Clerk

Probability Questions for SBI Clerk

Probability Questions for SBI Clerk 2015 Exam

Study the following information carefully to answer the questions that follow. A box contains 2 blue caps, 4 red caps, 5 green caps and 1 yellow cap:    (SBI Clerk 2015)

1. If four caps are picked at random, what is the probability that none is green ?
(a) \(\frac { 7 } { 99 }\)
(b) \(\frac { 5 } { 99 }\)
(c) \(\frac { 7 } { 12 }\)
(d) \(\frac { 5 } { 12 }\)
(e) None of these

Quantitative Aptitude

2. If three caps are picked at random, what is the probability that two are red and one is green ?
(a) \(\frac { 9 } { 22 }\)
(b) \(\frac { 6 } { 19 }\)
(c) \(\frac { 1 } { 6 }\)
(d) \(\frac { 3 } { 22 }\)
(e) None of these

3. If one cap is picked at random, what is the probability that it is either blue or yellow?
(a) \(\frac { 2 } { 9 }\)
(b) \(\frac { 1 } { 4 }\)
(c) \(\frac { 3 } { 8 }\)
(d) \(\frac { 6 } { 11 }\)
(e) None of these

Probability Questions for SBI Clerk 2015 Answers

1. (a) Total number of caps = 12
Total result n(S) \(= 12 C _ { 4 }\)
n(S) \(= \frac { 12 ! } { 4 ! \times 12 ! – 4 } = \frac { 12 \times 11 \times 10 \times 9 \times 8 ! } { 4 \times 3 \times 2 \times 1 \times 8 ! } = 5 \times 99\)
n(E₁) = Out of 5 caps, number of ways to not pick a green cap \(= ^ { 5 } \mathrm { C } _ { 0 }\)
n(E2) = Out of 7 caps, number of ways to not pick up 4 caps
\(= ^ { 7 } \mathrm { C } _ { 4 } = \frac { 7 ! } { 4 \times 7 ! – 4 } = \frac { 7 \times 6 \times 5 \times 4 \times 3 ! } { 4 \times 3 \times 2 \times 1 \times 3 ! } = 35\)
p(E) \(= \frac { n \left( E _ { 1 } \right) n \left( E _ { 2 } \right) } { n ( s ) } = \frac { 1 \times 35 } { 5 \times 99 } = \frac { 7 } { 99 }\)

2. (b) Total number of caps = 12
n(S) \(= 12 \mathrm { C } _ { 3 } = \frac { 12 ! } { 3 ! \times 12 ! – 3 } = \frac { 12 \times 11 \times 10 \times 9 ! } { 3 \times 2 \times 1 \times 9 ! } = 220\)
n(E₁) = Out of 4 red caps, number of ways to pick 2 caps \(= ^ { 4 } \mathrm { C } _ { 2 }\)
\(= \frac { 4 ! } { 2 ! \times 4 ! – 2 } = \frac { 4 \times 3 \times 2 \times 1 } { 2 \times 1 \times 2 \times 1 } = 6\)
n(E₂) = Out of 5 green caps, number of ways to pick one cap \(= ^ { 5 } C _ { 1 } = 5\)
\(\mathrm { p } ( \mathrm { E } ) = \frac { \mathrm { n } \left( \mathrm { E } _ { 1 } \right) \times \mathrm { n } \left( \mathrm { E } _ { 2 } \right) } { \mathrm { n } ( \mathrm { S } ) } = \frac { 6 \times 5 } { 220 } = \frac { 3 } { 22 }\)

3. (b) Total number of caps = 12
n(S) \(= ^ { 12 } \mathrm { C } _ { 1 } = 12\)
Out of (2 blue + 1 yellow) caps number of ways to pick one cap n(E) \(= ^ { 3 } \mathrm { C } _ { 1 } = 3\)
Required probability  p(E) \(= \frac { n ( E ) } { n ( S ) } = \frac { 3 } { 12 } = \frac { 1 } { 4 }\)

Probability Questions for SBI Clerk 2012 Exam

1. In how many different ways can the letters of the word ‘CREAM’ be arranged ?
(a). 720
(b). 240
(c). 360
(d). 504
(e). None of these

Probability Questions for SBI Clerk 2012 Answers

1. (e)
C    R    E    A    M
1    2     3   4     5
Required number of ways = 5!
= 5 × 4 × 3 × 2 × 1 = 120