## Probability Questions for IBPS RRB

### Probability Questions for IBPS RRB 2017 Exam

**1. **In bag A there are 8 red balls, X green balls and 6 yellow balls. Probability of drawing one green from bag A is 3/5. In bag B are (X – 3) red balls, (X – 9) green balls and 6 yellow balls. 2 balls are drawn from bag B. Find the probability that both the balls are red colour ? **(IBPS RRB Scale-I Main 2017)**

(a). 12/63

(b). 13/70

(c). 14/75

(d). 17/70

(e). None of these

**2. **A bag contains 6 Red, 5 Green and 4 Yellow coloured balls. 2 balls are drawn at random after one another without replacement then what is the probability that atleast one ball is Green.

(a). \(\frac { 2 } { 3 }\)

(b). \(\frac { 4 } { 5 }\)

(c). \(\frac { 3 } { 8 }\)

(d). \(\frac { 4 } { 7 }\)

(e). \(\frac { 2 } { 7 }\)

**Probability Questions for IBPS RRB 2017 Answers**

**1. **(d) Probability of drawing one green ball

⇒ x = 21

⇒ Green balls = 21

In Bag B

Red balls = (x – 3) = (21 – 3) = 18

Green balls = (x – 9) = (21 – 9) = 12

Yellow balls = 6

⇒ Required probability

\(= \frac { 18 \mathrm { C } _ { 2 } } { 36 } = \frac { 18 \times 17 } { 36 \times 85 } = \frac { 17 } { 70 }\)

**2. **(d) According to question,

Probability that no ball is green

\(\frac { 10 \mathrm { C } _ { 1 } \times ^ { 9 } \mathrm { C } _ { 1 } } { 15 \times 14 } = \frac { 90 } { 15 \times 14 } = \frac { 3 } { 7 }\)

Required probability \(= 1 – \frac { 3 } { 7 } = \frac { 4 } { 7 }\)

### Probability Questions for IBPS RRB 2016 Exam

**1. **From a pack 52 cards, 3 cards are drawn together at random, What is the probability of both the cards are king ?

(a) \(\frac { 6 } { 7 }\)

(b) \(\frac { 7 } { 6 }\)

(c) \(\frac { 5 } { 6 }\)

(d) \(\frac { 11 } { 21 }\)

(e) \(\frac { 12 } { 21 }\)

**Probability Questions for IBPS RRB 2016 Answers**

**1. **(a)

\(^ { 7 } \mathrm { C } _ { 2 } = \frac { 7 \times 6 } { 2 } = 21\)

\(^ { 4 } \mathrm { C } _ { 1 } \times ^ { 3 } \mathrm { C } _ { 1 } + ^ { 4 } \mathrm { C } _ { 2 } = 4 \times 3 + \frac { 4 \times 3 } { 2 }\)

\(= 4 \times 3 + \frac { 4 \times 3 } { 2 } = 12 + 6 = 18\)

\(\mathrm { P } = \frac { 18 } { 21 } = \frac { 6 } { 7 }\)

### Probability Questions for IBPS RRB 2013 Exam

**Study the information carefully to answer the questions that follow.(IBPS RRB OS 2013)**

A bucket contains 8 red, 3 blue and 5 green marbles

**1. **If 4 marbles are drawn at random, what is the probability that 2 are red and 2 are blue ?

(a). \(\frac { 11 } { 16 }\)

(b). \(\frac { 3 } { 16 }\)

(c). \(\frac { 11 } { 72 }\)

(d). \(\frac { 3 } { 65 }\)

(e). None of these

**2. **If 2 marbles are drawn at random, what is the probability that both are green ?

(a). \(\frac { 1 } { 8 }\)

(b). \(\frac { 5 } { 16 }\)

(c). \(\frac { 2 } { 7 }\)

(d). \(\frac { 3 } { 8 }\)

(e). None of these

**3. **If 3 marbles are drawn at random. What is the probability that none is red ?

(a). \(\frac { 3 } { 8 }\)

(b). \(\frac { 1 } { 16 }\)

(c). \(\frac { 1 } { 10 }\)

(d). \(\frac { 3 } { 16 }\)

(e). None of these

**From the following, different committees are to be made as per the requirement given in each question.
In how many different ways can it be done ? (IBPS RRB OS 2013)**

10 men and 8 women out of which 5 men are teachers, 3 men doctors and businessmen. Among the women, 3 are teachers, 2 doctors, 2 researchers and 1 social worker.

**4. **A Committee of 5 in which 3 men and 2 women are there

(a). 3360

(b). 8568

(c). 4284

(d). 1680

(e). None of these

**5. **A Committee of 4 in which at least 2 women are there

(a). 1260

(b). 1820

(c). 3060

(d). 1890

(e). None of these

**6. **A Committee is 5 in which 2 men teachers, 2 women teachers and 1 doctor are there

(a). 75

(b). 150

(c). 214

(d). 20

(e). None of these

**7. **A Committee of 7.

(a). 31824

(b). 1200

(c). 9600

(d). 15912

(e). None of these

**8. **A Committee of 3 in which there is no teacher and no doctor

(a). 100

(b). 120

(c). 10

(d). 12

(e). None of these

**Probability Questions for IBPS RRB 2013 Answers**

**1.** (d) Total number of marbles = 8 + 3 + 5 = 16

n(S) = Exhaustive number of cases

= Number of ways of drawing 4 marbles out of 16

\(= 16 C _ { 4 } = \frac { 16 \times 15 \times 14 \times 13 } { 1 \times 2 \times 3 \times 4 } = 1820\)

n(E) = Number of cases when 2 marbles are red and 2 are blue.

\(= ^ { 8 } \mathrm { C } _ { 2 } \times ^ { 3 } \mathrm { C } _ { 2 } = \frac { 8 \times 7 } { 1 \times 2 } \times \frac { 3 \times 2 } { 1 \times 2 } = 84\)

∴ Required probability \(= \frac { 84 } { 1820 } = \frac { 3 } { 65 }\)

**2. **(e)

n(S) \(= 16 \mathrm { C } _ { 2 } = \frac { 16 \times 15 } { 1 \times 2 } = 120[latex]

n(E) [latex]= ^ { 5 } \mathrm { C } _ { 2 } = \frac { 5 \times 4 } { 1 \times 2 } = 10\)

Required probability \(= \frac { 10 } { 120 } = \frac { 1 } { 12 }\)

**3. **(c)

n(S) \(= 16 \mathrm { C } _ { 2 } = \frac { 16 \times 15 \times 14 } { 1 \times 2 \times 3 } = 560\)

Out of the three drawn marbles none is red.

Clearly they will be either blue or green.

\(\therefore \quad n ( \mathrm { E } ) = ^ { 8 } \mathrm { C } _ { 3 } = \frac { 8 \times 7 \times 6 } { 1 \times 2 \times 3 } = 56\)

\(\therefore \quad \text { Required probability } = \frac { n ( E ) } { n ( S ) } = \frac { 56 } { 560 } = \frac { 1 } { 10 }\)

**4. **(a)The committee consists of 3 men and 2 women.

Out of 10 men. 3 men can be selected in ^{10}C_{3} ways

and out of 8 woman can be selected in ^{8}C_{2} ways

∴ Total number of selections = ^{10}C_{3} × ^{8}C_{2}

\(= \frac { 10 \times 9 \times 8 } { 1 \times 2 \times 3 } \times \frac { 8 \times 7 } { 1 \times 2 }\)

= 3360

**5. **(d) Let’s look at following cases :

(i) 2 women & 2 women

(ii) 3 women & 1 man

(iii) 4 women

∴ Total number of selections = ^{8}C_{2} × ^{10}C_{2} + ^{8}C_{3} × ^{10}C_{1} + ^{8}C_{4}

\(= \frac { 8 \times 7 } { 1 \times 2 } \times \frac { 10 \times 9 } { 1 \times 2 } + \frac { 8 \times 7 \times 6 \times 10 } { 1 \times 2 \times 3 } + \frac { 8 \times 7 \times 6 \times 5 } { 1 \times 2 \times 3 \times 4 }\)

= 1260 + 560 + 70 = 1890

**6.** (b) Out of 5 men 2 teachers, can be selected in ^{5}C_{2} ways.

Out of 3 women teachers, 2 can be selected in ^{3}C_{2} ways.

Out of 5 doctors 1 can be selected in ^{5}C_{1} ways.

∴ Total number of selections = ^{5}C_{2} × ^{3}C_{2} × ^{5}C_{1}

\(= \frac { 5 \times 4 } { 1 \times 2 } \times \frac { 3 \times 2 } { 1 \times 2 } \times 5\)

= 10 × 3 × 5 = 150

**7.** (a) Out of 18 persons, a committee of 7 persons is to be formed.

Total number of selections = ^{18}C_{7}

\(= \frac { 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 } { 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 } = 31824\)

**8. **(c) The committee has no teachers and no doctor.

Out of 18 persons, there are 8 teacher and 5 doctors.

∴ Total number of selections =Number of ways of selecting 3 persons out of remaining 5 persons.

\(= ^ { 5 } C _ { 3 } = \frac { 5 \times 4 \times 3 } { 1 \times 2 \times 3 } = 10\)

## Leave a Reply