BANK & INSURANCE (PROBABILITY) PART 3

Total Questions: 45

11. Directions (11-12): Answer the questions based on the information given below

In a biased coin, the probability of getting a head is 4 times more than that of getting a tail. Two such biased coins were tossed simultaneously such that probability of getting head on one coin and tail on other is ‘x’. A bag contains 2 white balls, ‘y’ pink balls and 10 red balls. If two balls are picked at random from bag then the probability that both balls are pink is equal to (x – 157)/630.

Ques: Find the total number of balls in the bag

Correct Answer: (c) 15
Solution:

Let the probability of getting a tail be ‘a’
Therefore, probability of getting a head = a + 4a = 5a
Hence, a + 5a = 1
a = 1/6

Therefore, x = 2 × (1/6) × (5/6) = 5/18

Total number of balls in the bag = (2 + y + 10)
= (12 + y)

According to the question,
yC₂/(12 + y)C₂ = (x − 157)/630

y(y − 1)/(12 + y)(11 + y) = 5/18 − 157/630

y(y − 1)/(12 + y)(11 + y) = 1/35

35y² − 35y = y² + 23y + 132
34y² − 58y − 132 = 0
34y² − 102y + 44y − 132 = 0
34y(y − 3) + 44(y − 3) = 0
(y − 3)(34y + 44) = 0

Since the number of balls cannot be negative
therefore, y = 3

Total number of balls in the bag = 2 + 3 + 10 = 15
Hence, option c

12. Two balls are picked at random from the bag. Find the probability that both balls are red balls.

Correct Answer: (a) 3/7  
Solution:

Let the probability of getting a tail be ‘a’
Therefore, probability of getting a head = a + 4a = 5a
Hence, a + 5a = 1
a = 1/6

Therefore, x = 2 × (1/6) × (5/6) = 5/18

Total number of balls in the bag = (2 + y + 10)
= (12 + y)

According to the question,
yC₂/(12 + y)C₂ = (x − 157)/630

y(y − 1)/(12 + y)(11 + y) = 5/18 − 157/630

y(y − 1)/(12 + y)(11 + y) = 1/35

35y² − 35y = y² + 23y + 132
34y² − 58y − 132 = 0
34y² − 102y + 44y − 132 = 0
34y(y − 3) + 44(y − 3) = 0
(y − 3)(34y + 44) = 0

Since the number of balls cannot be negative
therefore, y = 3

Required probability = 10C₂/15C₂ = 3/7
Hence, option a.

13. Directions (13-14): Answer the questions based on the information given below.

A bag contains red and green balls, only in the ratio of 3:2, respectively. A box contains 7 red, 8 blue and 6 yellow balls. Two balls are randomly drawn from the box and put into the bag. Then a ball is taken out of the bag such that the probability that a red ball is drawn is 5/9.

Ques: Number of balls in the bag initially was:

Correct Answer: (d) 10  
Solution:

Let the number of red and green balls in the bag be
‘3x’ and ‘2x’, respectively.

Case I: Two red balls are drawn from the box and are put into the bag.
The probability of drawing a red ball from the bag
= (7C₂/21C₂) × ((3x + 2)/(5x + 2)) = (1/10) × ((3x + 2)/(5x + 2))

Case II: Only one red ball is drawn from the box and is put into the bag.
The probability of drawing a red ball from the bag
= (7C₁ × 14C₁/21C₂) × ((3x + 1)/(5x + 2))
= (7/15) × ((3x + 1)/(5x + 2))

Case III: No red ball is drawn from the box and put into the bag.
The probability of drawing a red ball from the bag
= (14C₂/21C₂) × (3x/(5x + 2)) = (13/30) × (3x)/(5x + 2)

So according to question:
(1/10) × ((3x + 2)/(5x + 2)) + (7/15) × ((3x + 1)/(5x + 2)) + (13/30) × (3x)/(5x + 2) = 5/9

[(3x + 2)/(10(5x + 2))] + [(7(3x + 1))/(15(5x + 2))] + [(39x)/(30(5x + 2))] = 5/9

3(90x + 20) = 50(5x + 2)
270x + 60 = 250x + 100
20x = 40
x = 2

So, number of red and green balls in the bag is ‘6’ and ‘4’, respectively.
Number of balls in the bag initially = 10
Hence, option d.

14. If all the balls in box are put in the bag and now two balls are randomly drawn from the bag, then what is the probability that a yellow ball and a blue ball are drawn?

Correct Answer: (b) 16/155  
Solution:

Let the number of red and green balls in the bag be
‘3x’ and ‘2x’, respectively.

Case I: Two red balls are drawn from the box and are put into the bag.
The probability of drawing a red ball from the bag
= (7C₂/21C₂) × ((3x + 2)/(5x + 2)) = (1/10) × ((3x + 2)/(5x + 2))

Case II: Only one red ball is drawn from the box and is put into the bag.
The probability of drawing a red ball from the bag
= (7C₁ × 14C₁/21C₂) × ((3x + 1)/(5x + 2))
= (7/15) × ((3x + 1)/(5x + 2))

Case III: No red ball is drawn from the box and put into the bag.
The probability of drawing a red ball from the bag
= (14C₂/21C₂) × (3x/(5x + 2)) = (13/30) × (3x)/(5x + 2)

So according to question:
(1/10) × ((3x + 2)/(5x + 2)) + (7/15) × ((3x + 1)/(5x + 2)) + (13/30) × (3x)/(5x + 2) = 5/9

[(3x + 2)/(10(5x + 2))] + [(7(3x + 1))/(15(5x + 2))] + [(39x)/(30(5x + 2))] = 5/9

3(90x + 20) = 50(5x + 2)
270x + 60 = 250x + 100
20x = 40
x = 2

So, number of red and green balls in the bag is ‘6’ and ‘4’, respectively.

Total number of balls in the bag now = 10 + 21 = 31

Desired probability = (8C₁ × 6C₁)/(31C₂)
= (8 × 6 × 2)/(31 × 30) = 16/155

Hence, option b.

15. Directions (15-16): Answer the questions based on the information given below.

A magician has different number of doves of three different colours (white, grey and purple) in a box. The number of white doves is 25% more than that of purple doves. Total number of doves in the box is 48. The number of purple doves is at least 10, and the number of white doves is less than grey doves.

Ques: If two doves are taken out from the box by the magician, then find the probability that both doves are of white colour.

Correct Answer: (b) 35/376  
Solution:

Let the number of purple doves be ‘x’.
Then, number of white doves = 1.25x
Let the number of grey doves = y
So, x + 1.25x + y = 48
2.25x + y = 48 ........(1)
Since, x ≥ 10
When x = 10,
Number of white doves will be in fraction which is not possible.
So, x = 11 also not possible.
When x = 12,
Number of white doves = 1.25 × 12 = 15
So, y = 48 − 2.25 × 12 = 21
Since, number of white doves is less than grey doves,
so this is possible.
Similarly, x = 13, 14 and 15 not possible because of fraction value.
When x = 16,
Number of white doves = 1.25 × 16 = 20
Number of grey doves, y = 48 − 2.25 × 16 = 12
So, this case is not possible.
Therefore, x = 12 and y = 21
So, number of purple doves = 12
Then, number of white doves = 15
Let the number of grey doves = 21
Required probability = 15C₂/48C₂ = 35/376
Hence, option b.

16. What is the probability of drawing a purple colour dove from a box?

Correct Answer: (c) 1/4
Solution:Let the number of purple doves be ‘x’.
Then, number of white doves = 1.25x
Let the number of grey doves = y
So, x + 1.25x + y = 48
2.25x + y = 48 ........(1)
Since, x ≥ 10
When x = 10,
Number of white doves will be in fraction which is not possible.
So, x = 11 also not possible.
When x = 12,
Number of white doves = 1.25 × 12 = 15
So, y = 48 − 2.25 × 12 = 21
Since, number of white doves is less than grey doves,
so this is possible.
Similarly, x = 13, 14 and 15 not possible because of fraction value.
When x = 16,
Number of white doves = 1.25 × 16 = 20
Number of grey doves, y = 48 − 2.25 × 16 = 12
So, this case is not possible.
Therefore, x = 12 and y = 21
So, number of purple doves = 12
Then, number of white doves = 15
Let the number of grey doves = 21
Required probability = 12/48 = 1/4
Hence, option c

17. Two persons Ram and Vijay started playing a dart throwing game such that Ram got the first chance to throw the dart. It is also known that Vijay will get the chance to throw the dart only when Ram misses his target. The probability that Ram and Vijay will throw the dart on target is 0.4 and 0.2, respectively. Find the probability that Vijay will hit the target.

Correct Answer: (c) 3/13
Solution:

Probability that Ram will hit the target, P(R) = 0.4
Probability that Vijay will hit the target, P(V) = 0.2
Probability that Ram will miss the target, P(R’)
= 1 − 0.4 = 0.6

Probability that Vijay will miss the target, P(V’) = 1 − 0.2 = 0.8

Required probability = P(R’) × P(V) + P(R’) × P(V’) × P(R’) × P(V) + .....

Required probability = 0.6 × 0.2 + 0.6 × 0.8 × 0.6 × 0.2 + .....

Required probability = 0.12{1 + 0.48 + 0.48² + ......}

The above expression in the bracket is a geometric progression where
1st term (a) = 1, common ratio (r) = (0.48/1) = 0.48

Therefore, using sum of infinite terms of a geometric progression = [a/(1 − r)] where r < 1

Required probability = 0.12[1/(1 − 0.48)]
Required probability = 0.12/0.52 = 3/13
Hence, option c.

18. A bag contains 4 red balls, 6 blue balls and 5 green balls. Ashish picks 2 balls at random without replacement such that the probability that he claims that both the balls are of same colour is (97/280). If it is known that Ashish does not always speak the truth, then find the probability of Ashish speaking the truth.

Correct Answer: (d) (7/8)  
Solution:

Let the probability that Ashish speak the truth = ‘x’
Then, probability that Ashish lies = (1 − x)

Probability of picking two balls of the same colour
= (4C₂ + 6C₂ + 5C₂) ÷ 15C₂ = (6 + 15 + 10)/105
= (31/105)

Probability of picking two balls not of the same colour
= [15C₂ − (4C₂ + 6C₂ + 5C₂)] ÷ 15C₂
= (74/105)

Probability that Ashish claims he picked two balls of the same colour
= Probability that Ashish picks two balls of the same colour × probability that Ashish speaks the truth + Probability that Ashish picks two balls of different colour × Probability that Ashish lies

= (31/105) × x + (74/105) × (1 − x)

(31x/105) + (74/105 − 74x/105) = (97/280)

(74 − 43x)/105 = (97/280)

10185 = 20720 − 12040x

x = 10535 ÷ 12040 = 0.875

x = (7/8)

Hence, option d.

19. A dice is biased in such a way that if thrown once, then the probability of getting an odd number on the dice is (1/9) more than the probability of getting an even number. Also, the probability of getting a particular even/odd number is same. If two such biased dice are thrown at once, then find the probability of getting a sum greater than or equal to 9.

Correct Answer: (b) (193/729)
Solution:

Let the probability of getting an even number = ‘x’
Then, probability of getting an odd number = (x + 1/9)

So, x + (x + 1/9) = 2x + (1/9) = 1

x = {1 − (1/9)} ÷ 2 = (4/9)

So, probability of getting an even number and an odd number on the die are (4/9) and (5/9), respectively Since there are 3 odd numbers and 3 even numbers on the die,

Probability of getting a specific even number
= (4/9) × (1/3) = (4/27)

Similarly, probability of getting a specific odd number
= (5/27)

On throwing two dice at once, all possible events where the sum is greater than or equal to 9 are
= {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5) and (6, 6)}

So, required probability = 3 × (4/27) × (4/27) + 6 × (5/27) × (4/27) + (5/27) × (5/27)

= (193/729)

Hence, option b.

20. A bag contains 12 red balls and 8 green balls while a box contains 8 red balls, 10 green balls and 7 white balls. A ball from bag is transferred in the box then three balls are randomly drawn from box. What will be the probability that exactly 2 balls are red in colour?

Correct Answer: (d) 711/3250  
Solution:

Case I: A red ball is transferred from bag to box.
Probability = 12C₁/20C₁ × 9C₁ × (7C₁)/26C₃ = 12/20 × (36 × 17)/2600 = 459/3250

Case II: A green ball is transferred from bag to box.
Probability = 8C₁/20C₁ × (8C₂ × 18C₁)/26C₃ = 8/20 × (28 × 18)/2600 = 252/3250

Required probability = 459/3250 + 252/3250
= 711/3250

Hence, option d