Probability Questions for IBPS RRB

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

Quantitative Aptitude

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 ¹⁰C₃ ways
and out of 8 woman can be selected in ⁸C₂ ways
∴ Total number of selections = ¹⁰C₃ × ⁸C₂
\(= \frac { 10 \times 9 \times 8 } { 1 \times 2 \times 3 } \times \frac { 8 \times 7 } { 1 \times 2 }\)
= 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 = ⁸C₂ × ¹⁰C₂ + ⁸C₃ × ¹⁰C₁ + ⁸C₄
\(= \frac { 8 \times 7 } { 1 \times 2 } \times \frac { 10 \times 9 } { 1 \times 2 } + \frac { 8 \times 7 \times 6 \times 10 } { 1 \times 2 \times 3 } + \frac { 8 \times 7 \times 6 \times 5 } { 1 \times 2 \times 3 \times 4 }\)
= 1260 + 560 + 70 = 1890

6. (b) Out of 5 men 2 teachers, can be selected in ⁵C₂ ways.
Out of 3 women teachers, 2 can be selected in ³C₂ ways.
Out of 5 doctors 1 can be selected in ⁵C₁ ways.
∴ Total number of selections = ⁵C₂ × ³C₂ × ⁵C₁
\(= \frac { 5 \times 4 } { 1 \times 2 } \times \frac { 3 \times 2 } { 1 \times 2 } \times 5\)
= 10 × 3 × 5 = 150

7. (a) Out of 18 persons, a committee of 7 persons is to be formed.
Total number of selections = ¹⁸C₇
\(= \frac { 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 } { 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 } = 31824\)

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