BANK & INSURANCE (PROBABILITY) PART 3

Total Questions: 45

31. Directions (31-33): Answer the questions based on the information given below.

Three bags A, B and C contains balls of three different colours i.e. red, blue and green. Bag A contains ‘x’ red, ‘x + 10’ green and ‘x + 4’ blue balls. If two balls are drawn from the bag at a random then the probability of getting a red and a green coloured ball together from bag ‘A’ is 1/5. Another bag B contains (x - y) red, (x + 3y) green and (2x - 3y) blue balls. If two balls are drawn from this bag at a random then the probability of getting a red and a green coloured ball together from bag ‘B’ is also 1/5. Total number of blue balls in bag A, B and C together is 20 and number of green balls in bag B is 62.5% more than number of green balls in bag C and the probability of drawing a red ball from bag ‘C’ is 2/5.

Ques: If all the balls in bag A, bag B and bag C is mixed together and are put in bag D, then find the probability of getting a red and a green ball, when two balls are drawn from bag at a random.

Correct Answer: (b) 6/25  
Solution:

Total number of balls in bag A = x + x + 10 + x + 4
= ‘3x + 14’

According to question:
{C₁ × (x + 10)C₁}/(3x + 14)C₂ = 1/5

Or, {2 × x × x × (x + 10)}/((3x + 14)(3x + 13)) = 1/5

Or, 10x² + 100x = 9x² + 81x + 182
Or, x² + 19x − 182 = 0
Or, x² + 26x − 7x − 182 = 0
Or, x(x + 26) − 7(x + 26) = 0
Or, (x − 7)(x + 26) = 0

Or, x = 7 or x = −26 (not possible)

So, number of red balls in bag A = 7
Number of green balls in bag A = 7 + 10 = 17
Number of green balls in bag A = 7 + 4 = 11

Total number of balls in bag A = 3 × 7 + 14 = 35

Number of red balls in bag B = x − y = (7 − y)
Number of green balls in bag B = x + 3y = (7 + 3y)
Number of blue balls in bag B = 2x − 3y
= (14 − 3y)

Total number of balls in bag B = 7 − y + 7 + 3y + 14 − 3y = (28 − y)

According to question:
{7C₁ × (7 + 3y)C₁}/(28 − y)C₂ = 1/5

Or, 10(7 − y)(7 + 3y) = (28 − y)(27 − y)

Or, 490 − 30y² + 140y = 756 + y² − 55y
Or, 31y² − 195y + 266 = 0
Or, 31y² − 62y − 133y + 166 = 0
Or, 31y(y − 2) − 133(y − 2) = 0
Or, (31y − 133)(y − 2) = 0

So, y = 2 or y = 133/31 (not possible)

Number of red balls in bag B = (7 − y) = 7 − 2 = 5
Number of green balls in bag B = (7 + 3y) = 7 + 6 = 13
Number of blue balls in bag B = (14 − 3y) = 14 − 6 = 8

Total number of balls in bag B = (28 − y) = 28 − 2 = 26

Number of blue balls in bag C = 20 − 11 − 8 = 1
Number of green balls in bag C = 13/1.625 = 8

Let number of red balls in bag C is ‘z’
So, z/(z + 9) = 2/5
Or, 5z = 2z + 18
Or, z = 6

Number of red balls in bag C = 6

Total number of balls in bag C = 1 + 8 + 6 = 15

Bags   Red balls   Green balls   Blue balls   Total number of balls
A      7           17            11           35
B      5           13            8            26
C      6           8             1            15

Total number of red balls in bag D = 7 + 5 + 6 = 18
Total number of green balls in bag D = 17 + 13 + 8 = 38
Total number of blue balls in bag D = 20

Total number of balls in bag D = 18 + 38 + 20 = 76

Desired probability = (¹⁸C₁ × ³⁸C₁)/⁷⁶C₂ = (2 × 18 × 38)/(76 × 75) = 6/25

Hence, option b.

32. Find the value of ‘x + y’.

Correct Answer: (e) 9
Solution:

Total number of balls in bag A = x + x + 10 + x + 4
= ‘3x + 14’

According to question;
{C₁ × (x + 10)C₁}/(3x + 14)C₂ = 1/5

Or, {2 × x × x × (x + 10)}/((3x + 14)(3x + 13)) = 1/5

Or, 10x² + 100x = 9x² + 81x + 182
Or, x² + 19x − 182 = 0
Or, x² + 26x − 7x − 182 = 0
Or, x(x + 26) − 7(x + 26) = 0
Or, (x − 7)(x + 26) = 0

Or, x = 7 or x = −26 (not possible)

So, number of red balls in bag A = 7
Number of green balls in bag A = 7 + 10 = 17
Number of green balls in bag A = 7 + 4 = 11

Total number of balls in bag A = 3 × 7 + 14 = 35

Number of red balls in bag B = x − y = (7 − y)
Number of green balls in bag B = x + 3y = (7 + 3y)
Number of blue balls in bag B = 2x − 3y
= (14 − 3y)

Total number of balls in bag B = 7 − y + 7 + 3y + 14
− 3y = (28 − y)

According to question;
{(7 − y)C₁ × (7 + 3y)C₁}/(28 − y)C₂ = 1/5

Or, 2 × (7 − y) × (7 + 3y)/((28 − y)(27 − y)) = 1/5

Or, 10(7 − y)(7 + 3y) = (28 − y)(27 − y)

Or, 490 − 30y² + 140y = 756 + y² − 55y
Or, 31y² − 195y + 266 = 0
Or, 31y² − 62y − 133y + 166 = 0
Or, 31y(y − 2) − 133(y − 2) = 0
Or, (31y − 133)(y − 2) = 0

So, y = 2 or y = 133/31 (not possible)

Number of red balls in bag B = (7 − y) = 7 − 2 = 5
Number of green balls in bag B = (7 + 3y) = 7 + 6 = 13
Number of blue balls in bag B = (14 − 3y) = 14 − 6 = 8

Total number of balls in bag B = (28 − y) = 28 − 2 = 26

Number of blue balls in bag C = 20 − 11 − 8 = 1
Number of green balls in bag C = 13/1.625 = 8

Let number of red balls in bag C is ‘z’
So, z/(z + 9) = 2/5
Or, 5z = 2z + 18
Or, z = 6

Number of red balls in bag C = 6

Total number of balls in bag C = 1 + 8 + 6 = 15

Bags   Red balls   Green balls   Blue balls   Total number of balls
A      7           17            11           35
B      5           13            8            26
C      6           8             1            15

x + y = 7 + 2 = 9

Hence, option e.

33. If three balls are drawn from bag C, then find the probability of getting different coloured balls.

Correct Answer: (b) 48/455  
Solution:

Total number of balls in bag A
= x + x + 10 + x + 4 = ‘3x + 14’

According to question;
{C₁ × (x + 10)C₁}/(3x + 14)C₂ = 1/5

Or, {2 × x × x × (x + 10)}/((3x + 14)(3x + 13)) = 1/5

Or, 10x² + 100x = 9x² + 81x + 182

(continued on right side)

Or, x² + 19x − 182 = 0
Or, x² + 26x − 7x − 182 = 0
Or, x(x + 26) − 7(x + 26) = 0
Or, (x − 7)(x + 26) = 0

Or, x = 7 or x = −26 (not possible)

So, number of red balls in bag A = 7
Number of green balls in bag A = 7 + 10 = 17
Number of green balls in bag A = 7 + 4 = 11

Total number of balls in bag A = 3 × 7 + 14 = 35

Number of red balls in bag B = x − y = (7 − y)
Number of green balls in bag B = x + 3y = (7 + 3y)
Number of blue balls in bag B = 2x − 3y
= (14 − 3y)

Total number of balls in bag B = 7 − y + 7 + 3y + 14
− 3y = (28 − y)

According to question;
{(7 − y)C₁ × (7 + 3y)C₁}/(28 − y)C₂ = 1/5

Or, 2 × (7 − y) × (7 + 3y)/((28 − y)(27 − y)) = 1/5

Or, 10(7 − y)(7 + 3y) = (28 − y)(27 − y)

Or, 490 − 30y² + 140y = 756 + y² − 55y
Or, 31y² − 195y + 266 = 0
Or, 31y² − 62y − 133y + 166 = 0
Or, 31y(y − 2) − 133(y − 2) = 0
Or, (31y − 133)(y − 2) = 0

So, y = 2 or y = 133/31 (not possible)

Number of red balls in bag B = (7 − y) = 7 − 2 = 5
Number of green balls in bag B = (7 + 3y) = 7 + 6 = 13
Number of blue balls in bag B = (14 − 3y) = 14 − 6 = 8

Total number of balls in bag B = (28 − y) = 28 − 2 = 26

Number of blue balls in bag C = 20 − 11 − 8 = 1
Number of green balls in bag C = 13/1.625 = 8

Let number of red balls in bag C is ‘z’
So, z/(z + 9) = 2/5
Or, 5z = 2z + 18
Or, z = 6

Number of red balls in bag C = 6

Total number of balls in bag C = 1 + 8 + 6 = 15

Bags   Red balls   Green balls   Blue balls   Total number of balls
A      7           17            11           35
B      5           13            8            26
C      6           8             1            15

Desired Probability = {6C₁ × 8C₁ × 1C₁}/15C₃
= (6 × 6 × 8 × 1)/(15 × 14 × 13) = 48/455

Hence, option b.

34. Directions (34-36): Answer the questions based on the information given below.

There are three bags i.e. ‘A’, ‘B’ and ‘C’ and each of them contains certain number of three coloured balls i.e. red, green and blue. Total number of balls in bag ‘B’ and bag ‘C’ is 36 and 48 respectively while total number of green balls in all three bags together is 29 and ratio of number of green balls in bag ‘B’ and bag ‘A’ is 3:4, respectively. Number of red balls in bag ‘B’ is 30% of total number of balls in bag ‘A’. Probability of drawing a red ball and a blue ball together (without replacement) from bag ‘A’ is (3/13) and that from bag ‘B’ is (2/7). Probability of drawing a blue ball from bag ‘C’ is 0.5. Number of red balls in bag ‘C’ is one more than number of blue balls in bag ‘B’.

Ques :Total number of red balls in all three bags together can be:

Correct Answer: (d) either (a) or (b)  
Solution:

Let number of green balls in bag ‘A’ and bag ‘B’ be ‘4x’ and ‘3x’, respectively.
Number of blue balls in bag ‘C’ = 0.50 × 48 = 24

Let number of red balls in bag ‘C’ be ‘y’.
So, number of blue balls in bag ‘B’ = y − 1
Number of green balls in bag ‘C’ = 29 − 7x

So, y + 29 − 7x = 24
Or, 7x − y = 5 .......... (1)

Number of red balls in bag ‘B’ = 36 − 3x − y + 1
= 37 − 3x − y

So, {2(37 − 3x − y)(y − 1)}/{36 × 35} = 2/7

Or, (37 − 3x − y)(y − 1) = 180

Or, (37 − 3x − y)(y − 1) = 180

Or, (37 − 3x − x + 5)(7x − 5 − 1) = 180 [from equation (1), y = 7x − 5]

Or, (42 − 10x)(7x − 6) = 180

Or, 294x − 70x² − 252 + 60x = 180

Or, 70x² − 354x + 432 = 0

Or, 70x² − 210x − 144x + 432 = 0

Or, 70x(x − 3) − 144(x − 3) = 0

Or, (70x − 144)(x − 3) = 0

Or, x = 3 or x = 144/70 (not possible)

And, y = 7 × 3 − 5 = 16

Number of red balls in bag ‘C’ = 16

Number of green balls in bag ‘C’ = 29 − 7 × 3 = 8

Number of green balls in bag ‘B’ = 3 × 3 = 9

Number of blue balls in bag ‘B’ = 16 − 1 = 15

Number of red balls in bag ‘B’ = 36 − 3 × 3 − 15 = 12

Number of green balls in bag ‘A’ = 4 × 3 = 12

Total number of balls in bag ‘A’ = 12/0.30 = 40

Let number of red balls in bag ‘A’ be ‘z’

So, number of blue balls in bag ‘A’ = 40 − 12 − z
= 28 − z

So, {2 × z × (28 − z)}/{40 × 39} = 3/13

Or, z(28 − z) = 180

Or, z² − 28z + 180 = 0

Or, z² − 10z − 18z + 180 = 0

Or, z(z − 10) − 18(z − 10) = 0

Or, (z − 10)(z − 18) = 0

Or, z = 10 or z = 18

So, number of red balls in bag ‘A’ can be either 10 or 18.

And, number of blue balls in bag ‘A’ can be either 18 or 10.

Total number of red balls in all three bags together can be either 38 (10 + 12 + 16) or 46 (18 + 12 + 16)

Hence, option d.

35. If three balls are drawn from bag ‘B’ one after another and without replacement then the probability that all three balls are green in colour is:

Correct Answer: (c) 1/85
Solution:

Let number of green balls in bag ‘A’ and bag ‘B’ be ‘4x’ and ‘3x’, respectively.

Number of blue balls in bag ‘C’ = 0.50 × 48 = 24

Let number of red balls in bag ‘C’ be ‘y’.

So, number of blue balls in bag ‘B’ = y − 1

Number of green balls in bag ‘C’ = 29 − 7x

So, y + 29 − 7x = 24

Or, 7x − y = 5 .......... (1)

Number of red balls in bag ‘B’ = 36 − 3x − y + 1
= 37 − 3x − y

So, {2(37 − 3x − y)(y − 1)}/{36 × 35} = 2/7

Or, (37 − 3x − y)(y − 1) = 180

Or, (37 − 3x − 7x + 5)(7x − 5 − 1) = 180 [from equation (1), y = 7x − 5]

Or, (42 − 10x)(7x − 6) = 180

Or, 294x − 70x² − 252 + 60x = 180

Or, 70x² − 354x + 432 = 0

Or, 70x² − 210x − 144x + 432 = 0

Or, 70x(x − 3) − 144(x − 3) = 0

Or, (70x − 144)(x − 3) = 0

Or, x = 3 or x = 144/70 (not possible)

And, y = 7 × 3 − 5 = 16

Number of red balls in bag ‘C’ = 16

Number of green balls in bag ‘C’ = 29 − 7 × 3 = 8

Number of green balls in bag ‘B’ = 3 × 3 = 9

Number of blue balls in bag ‘B’ = 16 − 1 = 15

Number of red balls in bag ‘B’ = 36 − 3 × 3 − 15 = 12

Number of green balls in bag ‘A’ = 4 × 3 = 12

Total number of balls in bag ‘A’ = 12/0.30 = 40

Let number of red balls in bag ‘A’ be ‘z’

So, number of blue balls in bag ‘A’ = 40 − 12 − z
= 28 − z

36. Number of blue balls in bag ‘A’ is:

Correct Answer: (e) Either (a) or (c)
Solution:

Let number of green balls in bag ‘A’ and bag ‘B’ be ‘4x’ and ‘3x’, respectively.

Number of blue balls in bag ‘C’ = 0.50 × 48 = 24

Let number of red balls in bag ‘C’ be ‘y’.

So, number of blue balls in bag ‘B’ = y − 1

Number of green balls in bag ‘C’ = 29 − 7x

So, y + 29 − 7x = 24

Or, 7x − y = 5 .......... (1)

Number of red balls in bag ‘B’ = 36 − 3x − y + 1
= 37 − 3x − y

So, {2(37 − 3x − y)(y − 1)}/{36 × 35} = 2/7

Or, (37 − 3x − y)(y − 1) = 180

Or, (37 − 3x − 7x + 5)(7x − 5 − 1) = 180 [from equation (1), y = 7x − 5]

Or, (42 − 10x)(7x − 6) = 180

Or, 294x − 70x² − 252 + 60x = 180

Or, 70x² − 354x + 432 = 0

Or, 70x² − 210x − 144x + 432 = 0

Or, 70x(x − 3) − 144(x − 3) = 0

Or, (70x − 144)(x − 3) = 0

Or, x = 3 or x = 144/70 (not possible)

And, y = 7 × 3 − 5 = 16

Number of red balls in bag ‘C’ = 16

Number of green balls in bag ‘C’ = 29 − 7 × 3 = 8

Number of green balls in bag ‘B’ = 3 × 3 = 9

Number of blue balls in bag ‘B’ = 16 − 1 = 15

Number of red balls in bag ‘B’ = 36 − 3 × 3 − 15 = 12

Number of green balls in bag ‘A’ = 4 × 3 = 12

Total number of balls in bag ‘A’ = 12/0.30 = 40

Let number of red balls in bag ‘A’ be ‘z’

So, number of blue balls in bag ‘A’ = 40 − 12 − z
= 28 − z

So, {2 × z × (28 − z)}/{40 × 39} = 3/13
Or, z(28 − z) = 180
Or, z² − 28z + 180 = 0
Or, z² − 10z − 18z + 180 = 0
Or, z(z − 10) − 18(z − 10) = 0
Or, (z − 10)(z − 18) = 0
Or, z = 10 or z = 18

So, number of red balls in bag ‘A’ can be either 10 or 18.
And, number of blue balls in bag ‘A’ can be either 18 or 10.
Number of blue balls in bag ‘A’ can be 10 or 18.
Hence, option e.

37. There are three types of candies ‘A’, ‘B’ and ‘C’ in a bag. The probability of picking a candy ‘C’ at random is 6/11. The total number of candy ‘B’ is half of ‘C’. The probability of picking 1 candy of type ‘A’ and 1 candy of type ‘B’ out of total number of candies is 72/715. Find the total number of candies in bag.

Correct Answer: (b) 66  
Solution:

Let total number of candies be 11x
Therefore, number of candies of ‘C’ type = 6x
Number of candies of ‘B’ type = 6x/2 = 3x
Number of candies of ‘A’ type = 2x

According to the question,
{(²C₁ × ³C₁)/(¹¹C₂)} = 72/715
Or, 2x × 3x / 11x(11x − 1) = 72/715
Or, 66x − 6 = 65x
Or, x = 6

Therefore, total number of candies = 11x = 66
Hence, option b.

38. A bag contains some balls of three different colours i.e., red, blue and green. The bag contains 20 red balls and the probability of picking a blue ball from the bag is (3/25) more than the probability of picking a green ball. If 40% balls in the bag are red, then find the number of blue balls in the bag.

Correct Answer: (d) 18  
Solution:

Total number of balls in the bag = 20 ÷ 0.40 = 50
Let the number of blue balls in the bag be ‘x’
So, number of green balls in the bag = 50 − 20 − x
= (30 − x)

ATQ:
(x/50) − {(30 − x)/50} = (3/25)
Or, x − (30 − x) = (3/25) × 50
Or, 2x − 30 = 6
Or, 2x = 36
So, x = 18

Therefore, the bag contains 18 blue balls.
Hence, option d.

39. There are 40 bulbs of either of the two colours i.e. red and blue in a bag given that three out of every four bulbs in the bag is red. If two balls are drawn from the bag at random and without replacement then find the probability that one of the balls is red and other is blue.

Correct Answer: (d) (5/13)  
Solution:Number of red bulbs in the bag = 40 × (3/4) = 30
Number of blue bulbs in the bag = 40 − 30 = 10
Required probability = (30/40) × (10/39) × 2
= (5/13)
Hence, option d.

40. A person has 3 pet dogs, 3 pet cats and 2 pet rabbits. He feeds only one pet animal at a time and feeds all pet animals of the same species before feeding the next species. In how many arrangements can he feed all his pet animals?

Correct Answer: (b) 432  
Solution:

Let all 3 dogs together form 1 unit, all 3 cats together form 1 unit and all 2 rabbits together form 1 unit
So, number of ways of arranging these 3 units = 3! ways

Within the 3 dogs, number of possible feeding arrangements = 3! ways
Similarly, within the 3 cats, number of possible feeding arrangements = 3! ways
Within the 2 rabbits, number of possible feeding arrangements = 2! ways

So, total possible feeding arrangement of all 8 pet animals = 3! × 3! × 3! × 2!
= 6 × 6 × 6 × 2 = 432 ways

Hence, option b.