## Probability Questions for SBI PO

### Probability Questions for SBI PO 2017 Exam

**1.** There are 27 cards having number 1 to 27. Two cards are picked at random one by one. What is the probability that sum of number on these 2 cards is odd? **(SBI PO Prelim Exam 2017)
**[latex]\left( a \right) \frac { 13}{ 27}[/latex]

[latex]\left( b \right) \frac {8 }{ 13}[/latex]

[latex]\left( c \right) \frac { 182}{ 729}[/latex]

[latex]\left( d \right) \frac { 14}{ 27}[/latex]

(e) None of these

**There are three bags A, B and C. In each bag there are three types of colored balls Yellow, Green and Black. (SBI PO Main Exam 2017)**

In bag A, no. of yellow colored balls are y and no. of green colored balls are g. Number of green colored balls are 4 more than the number of yellow colored balls. When one ball is picked at random then the probability of getting color ball is [latex]\frac { 5 }{ 13 } [/latex]. The value of y is [latex]18\frac { 2 }{ 11 } %[/latex] less than g.

In bag b, number of yellow colored balls is [latex]22\frac { 2 }{ 9 } %[/latex] more than that of bag A. If two balls are picked at random from bag B then the probability of getting both green color ball is [latex]\frac { 4 }{ 7 } [/latex]. Total number of ball in bag B is 75.

In bag C, the ratio of number of green colored balls and number of black colored balls is 7 : 5. Total number of green and black colored balls is 36. If one ball is picked at random then the probability of getting one yellow ball is [latex]\frac { 7 }{ 13} [/latex]

**2.** If x number of yellow balls from bag B are taken and placed into bag A and 20 % of black balls from the bag A are taken and placed int o in bag B. If we pick one ball from bag B then the probability that the ball is of balck color is [latex]\frac { 11 }{ 26} [/latex]. Then find the value of x?

(a) 5

(b) 6

(c) 3

(d) 2

(e) None of these

**3.** If one ball picked at random from each of the bag A and bag B then find the probability that both of the balls are of the same color?

[latex]\left( a \right) \frac { 21\times 47 }{ 65\times 75 } [/latex]

[latex]\left( b \right) \frac { 22\times 43}{ 65\times 75 } [/latex]

[latex]\left( c \right) \frac { 11\times 17}{ 65\times 75 } [/latex]

(d) Can’t be determined

(e) None of these

**4.** Difference between the number of green balls in bag A and bag C is how much percent more/less than the sum of the number of black balls in bag A and bag C together?

(a) 100%

(b) 95%

(c) 67.5%

(d) 102.5%

(e) None of these

**Probability Questions for SBI PO 2017 Answers**

**1.** (d)

Odd sum is there when one card drawn is odd and another even.

⇒ Required probability

[latex]=\left( \frac { 13 }{ 27 } \times \frac { 14 }{ 26 } \right) \left( \frac { 14 }{ 27 } \times \frac { 13 }{ 26 } \right) =\frac { 14 }{ 27 } [/latex]

**2.** (d)

After replacement →

Yellow no. of balls in bag B = 22 – x

Black no. of balls in bag B = 28 + 5 = 33

Green no. of balls in bag B = 25

Then,

[latex s=1]\frac { 33 } { 22 \therefore x + 33 + 25 } = \frac { 11 } { 26 }[/latex]

[latex s=1]\frac { 33 } { 80 – x } = \frac { 11 } { 26 }[/latex]

78 = 80 – x

x = 2

**3.** (e)

Required probability

[latex]= \frac { 18 } { 65 } \times \frac { 22 } { 75 } = \frac { 22 } { 65 } \times \frac { 25 } { 75 } = \frac { 25 } { 65 } \times \frac { 28 } { 75 }[/latex]

[latex]= \frac { 1646 } { 65 \times 75 }[/latex]

**4.** (c)

Required% [latex s=1]= \frac { 40 \therefore 1 } { 40 } \times 100[/latex]

[latex]= \frac { 39 } { 40 } \times 100[/latex]

= 97.5%

### Probability Questions for SBI PO 2013 Exam

**1.** A bag contains 7 blue balls and 5 yellow balls. If two balls are selected at random, what is the probability that none is yellow?** (SBI PO 2013)**

[latex]\left( a \right) \frac { 5 }{ 35 }[/latex]

[latex]\left( b \right) \frac { 5 }{ 22 }[/latex]

[latex]\left( c \right) \frac { 7 }{ 22 }[/latex]

[latex]\left( d \right) \frac { 7 }{ 33 }[/latex]

[latex]\left(e \right) \frac { 7 }{ 66 }[/latex]

**2.** A die is thrown twice. What is the probability of getting a sum 7 from both the throws?** (SBI PO 2013)**

[latex]\left( a \right) \frac { 5 }{ 18 }[/latex]

[latex]\left( b \right) \frac { 1 }{ 18 }[/latex]

[latex]\left( c \right) \frac { 1 }{ 9 }[/latex]

[latex]\left( d \right) \frac { 1 }{ 6 }[/latex]

[latex]\left( e \right) \frac { 5 }{ 36 }[/latex]

**Probability Questions for SBI PO 2013 Answers**

**1.** (c)

Blue balls = 7

None-ball out of two yellow

Yellow balls = 5

∴ Both balls are blue

Total balls = 12

∴P(bothblueballs)

[latex]=\frac { 7 }{ 12 } \times \frac { 6 }{ 11 } =\frac { 7 }{ 22 } [/latex]

**2.** (d)

Total possible outcomes when A die is tin-own twice = 36

Outcome for getting a sum 7 from both throwns = 6{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

[latex]\therefore \quad \mathrm { P } ( \mathrm { E } ) = \frac { 6 } { 36 } = \frac { 1 } { 6 }[/latex]

### Probability Questions for SBI PO 2011 Exam

**1.** Out of five girls and three boys, four children are to be randomly selected for a quiz contest. What is the probability that all the selected children are girls? **(SBI Associates PO 2011)**

[latex]\left( a \right) \frac { 1 }{ 14 }[/latex]

[latex]\left( b \right) \frac { 1 }{ 7 }[/latex]

[latex]\left( c \right) \frac { 5 }{ 17 }[/latex]

[latex]\left( d \right) \frac { 2 }{ 17 }[/latex]

(e) None of these

**Probability Questions for SBI PO 2011 Answer**

**1.** (a)

Total number of ways of selecting 4 children out of 8

[latex]= ^ { 8 } \mathrm { C } _ { 4 } = \frac { 8 \times 7 \times 6 \times 5 } { 1 \times 2 \times 3 \times 4 } = 70[/latex]

Number of ways of selecting 4 girls out of 5 = ^{5}C_{4} = 5

Required probability

[latex]= \frac { 5 } { 70 } = \frac { 1 } { 14 }[/latex]

### Probability Questions for SBI PO 2010 Exam

**Study the given information carefully and answer the questions that follow :**

**Anurn contains 6 red, 4 blue, 2 green and 3 yellow marbles.** **(SBI Rural Bus. PO 2010)**

**1.** If four marbles are picked at random, what is the probability that at least one is blue?

[latex]\left( a \right) \frac { 4 }{ 15 }[/latex]

[latex]\left( b \right) \frac { 69 }{ 91 }[/latex]

[latex]\left( c \right) \frac { 11 }{ 15 }[/latex]

[latex]\left( d \right) \frac { 22 }{ 91 }[/latex]

(e) None of these

**2.** If two marbles are picked at random, what is the probability that both are red?

[latex]\left( a \right) \frac { 1 }{ 6 }[/latex]

[latex]\left( b \right) \frac { 1 }{ 3 }[/latex]

[latex]\left( c \right) \frac { 2 }{ 15 }[/latex]

[latex]\left( d \right) \frac { 2 }{ 5 }[/latex]

(e) None of these

**3.** If three marbles are picked at random, what is the probability that two are blue and one is yellow?

[latex]\left( a \right) \frac { 3 }{ 31 }[/latex]

[latex]\left( b \right) \frac { 1 }{ 5 }[/latex]

[latex]\left( c \right) \frac { 18 }{ 455 }[/latex]

[latex]\left( d \right) \frac { 7 }{ 15 }[/latex]

(e) None of these

**4.** If four marbles are picked at random, what is the probability that one is green, two are blue and one is red?

[latex]\left( a \right) \frac { 24 }{ 455 }[/latex]

[latex]\left( b \right) \frac { 13 }{ 35 }[/latex]

[latex]\left( c \right) \frac { 11 }{ 15 }[/latex]

[latex]\left( d \right) \frac { 1 }{ 3 }[/latex]

(e) None of these

**5.** If two marbles are picked at random, what is the probability that either both are green or both are yellow?

[latex]\left( a \right) \frac { 5}{ 91 }[/latex]

[latex]\left( b \right) \frac { 1 }{ 35 }[/latex]

[latex]\left( c \right) \frac { 1 }{ 3 }[/latex]

[latex]\left( d \right) \frac { 4 }{ 105 }[/latex]

(e) None of these

**A basket contains 4 red, 5 blue and 3 green marbles. (SBI Associates Bank PO 2010)**

**6.** If three marbles are picked at random, what is the probability that either all are green or all are red ?

[latex]\left( a \right) \frac { 7 }{ 44 }[/latex]

[latex]\left( b \right) \frac { 7 }{ 12 }[/latex]

[latex]\left( c \right) \frac { 5 }{ 12 }[/latex]

[latex]\left( d \right) \frac { 1 }{ 44 }[/latex]

(e) None of these

**7.** If two marbles are picked at random, What is the probability that both are red ?

[latex]\left( a \right) \frac { 3 }{ 7 }[/latex]

[latex]\left( b \right) \frac { 1 }{ 2 }[/latex]

[latex]\left( c \right) \frac { 2 }{ 11 }[/latex]

[latex]\left( d \right) \frac { 1 }{ 6 }[/latex]

(d) None of these

**8.** If three marbles are picked at random, What is the probability that at least one is blue?

[latex]\left( a \right) \frac { 7 }{ 12 }[/latex]

[latex]\left( b\right) \frac { 37 }{ 44 }[/latex]

[latex]\left( c \right) \frac { 5 }{ 12 }[/latex]

[latex]\left( d \right) \frac { 7 }{ 44 }[/latex]

(e) None of these

**A committee of five members is to be formed out of 3 trainees, 4 professors and 6 research associates. In how many different ways can this be done if: (SBI Associates Bank PO 2010)**

**9.** The committee should have all 4 professors and 1 research associate or all 3 trainees and 2 professors ?

(a) 12

(b) 13

(c) 24

(d) 52

(e) None of these

**10.** The committee should have 2 trainees and 3 research associates ?

(a) 15

(b) 45

(c) 60

(d) 9

(e) None of these

**Probability Questions for SBI PO 2010 Answers**

**1.** (b)

Number of way [latex]\frac { 5.72 } { 3 }[/latex] lakh of selecting 4 marbles out of 15 marbles

[latex]= ^ { 15 } \mathrm { C } _ { 4 } = \frac { 15 \times 14 \times 13 \times 12 } { 4 \times 3 \times 2 \times 1 } = 1365][/latex]

Number of ways of selecting 4 marbles when no one is blue

[latex]= ^ { 11 } \mathrm { C } _ { 4 } = \frac { 11 \times 10 \times 9 \times 8 } { 4 \times 3 \times 2 \times 1 } = 330[/latex]

Probability of getting 4 marble (when no one is blue)

= \frac { 330 } { 1365 } = \frac { 22 } { 91 }

[latex]Probability that at least one is blue [/latex]

[latex]= 1 – \frac { 22 } { 91 } = \frac { 69 } { 91 }[/latex]

**2.** (e)

Number of ways of selecting 2 red marbles from 6 red marbles = ^{6}C_{2}= 15

Number of ways of selecting 2 marbles from urn = ^{15}C_{2}=105

Required Probability

[latex]= \frac { 15 } { 105 } = \frac { 1 } { 7 }[/latex]

**3.** (c)

Number of ways of selecting 2 blue and one yellow marble = ^{4}C_{2} × ^{3}C_{1}= 6 × 3 = 18

Number of ways of selecting 3 marble from urn = ^{15}C_{3} = 455

Required Probability

[latex]= \frac { 18 } { 455 }[/latex]

**4.** (a)

Number of ways of selecting one green, two blue and one red marble = ^{2}C_{1} × ^{4}C_{2} × ^{6}C_{1}.

= 2 × 6 × 6 = 72

Number of ways of selecting 4 marbles from urn = ^{15}C_{4}

[latex]= \frac { 12 \times 13 \times 14 \times 15 } { 4 \times 3 \times 2 \times 1 } = 1365[/latex]

Required Probability

[latex]= \frac { 72 } { 1365 } = \frac { 24 } { 455 }[/latex]

**5.** (d)

Number of ways of selecting either two green marbles or two yellow marbles = ^{2}C_{2} + ^{3}C_{2} = 1 + 3 = 4

Number of ways of selecting 2 marbles = ^{15}C_{2} = 105

Required Probability

[latex]= \frac { 4 } { 105 }[/latex]

**6.** (d)

Total possible outcomes = Number of ways of picking

3 marbles out of 12 marbles = n(S)

[latex]= 12 _ { \mathrm { c3 } } \frac { 12 \times 11 \times 10 } { 1 \times 2 \times 3 } = 220[/latex]

Favourable number of cases = n(E)

= ^{3}C_{3} + ^{4}C_{3}

= 1 + 4=5

∴ Required probability

[latex]= \frac { \mathrm { n } ( \mathrm { E } ) } { \mathrm { n } ( \mathrm { S } ) } = \frac { 5 } { 220 } = \frac { 1 } { 44 }[/latex]

**7.** (e)

Total possible outcomes

[latex]= \mathrm { n } ( \mathrm { S } ) = ^ { 12 } \mathrm { C } _ { 2 } = \frac { 12 \times 11 } { 1 \times 2 } = 66[/latex]

Favourable number of cases = n(E)

[latex]= ^ { 4 } \mathrm { C } _ { 2 } = \frac { 4 \times 3 } { 1 \times 2 } = 6[/latex]

∴ Required probability

[latex]= \frac { \mathrm { n } ( \mathrm { E } ) } { \mathrm { n } ( \mathrm { S } ) } = \frac { 6 } { 66 } = \frac { 1 } { 11 }[/latex]

**8.** (b)

Total possible outcomes = n(S) = ^{12}C_{3} = 220

Favourable number of ways of picking 3 marbles (none is blue) out of 7 marbles

[latex]= ^ { 7 } \mathrm { C } _ { 3 } = \frac { 7 \times 6 \times 5 } { 1 \times 2 \times 3 } = 35[/latex]

∴ Required probability

[latex]= \left( 1 – \frac { 35 } { 220 } \right) = 1 – \frac { 7 } { 44 } = \frac { 37 } { 44 }[/latex]

**9.** (a)

Number of combinations

= (^{4}C_{4} × ^{6}C_{1} + ^{3}C_{3} × ^{4}C_{2}) = 1 × 6 + 1 × 6 = 12

**10.** (c)

Number of combinations

= Selecting 2 trainees out of 3 and selecting 3 research associates out of 6 = ^{3}C_{2} × ^{6}C_{3}

[latex]= 3 \times \frac { 6 \times 5 \times 4 } { 1 \times 2 \times 3 } = 60[/latex]

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