BANK & INSURANCE (PROBABILITY) PART 2

Total Questions: 45

1. A goldsmith has a bag, which contains some colourful stones. It contains 4 White and 8 Black stone. There is another bag which contains 5 White and 5 Black stones. One stone is to drawn from either of the two bags. What is the probability of drawing a White stone?

Correct Answer: (c) 5/12
Solution:CALCULATION :
Probability of choosing one bag = 1/2
Probability of White stone from 1st bag
=1/2× ⁴C₁ ¹²C₁=4/24=1/6
Probability of White stones from 2nd bag
= 1/2 × ⁵C₁/ ¹⁰C₁ = 5/20 = 1/4
∴ Required probability = 1/6 + 1/4 = 5/12

2. A TCS placement session is going on. There are 25 CS engineers and 5 EE engineers in the waiting hall. If three engineers are selected at random, the probability that 1 CS engineer and 2 EE engineers are selected is:

Correct Answer: (a) 25/406  
Solution:

Let S be the sample space.
Total no. of ways of selecting 3 engineers from 30 engineers = n(S) = ³⁰C₃
Let E be the event of selecting 1 CS and 2 EE engineers
⇒ n(E) = no. of ways of selecting 1 CS and 2 EE engineers

∴ The no. of ways in which 1 CS engineer from 25 and 2 EE engineers from 5 can be selected
= ²⁵C₁ × ⁵C₂
⇒ n(E) = ²⁵C₁ × ⁵C₂
⇒ P(E) = n(E)/n(S)
⇒ ²⁵C₁ × ⁵C₂ / ³⁰C₃
⇒ (25 × 10)/[(30 × 29 × 28)/(1 × 2 × 3)]
⇒ 250/4060
⇒ 25/406

∴ The probability that 1CS and 2 EE engineers are selected is 25/406.

3. A committee is to be formed in a college. Committee consists of 2 professors and 2 students. If there are 10 professors and 20 students. Find the number of ways to form a committee.

Correct Answer: (a) 8550  
Solution:Select 2 committee members from professors and 2 committee members from students
⇒ ¹⁰C₂ × ²⁰C₂
⇒ (10!)/(2! × 8!) [(20!)/(2! × 18!)]
⇒ [(10 × 9 × 8!)/(2! × 8!)] [(20 × 19 × 18!)/(2! × 18!)]
⇒ 45 × 190 = 8550 ways
∴ Total number of ways to form a committee is 8550

4. In a bag, there are 5 red, 7 white and 3 green balls. One ball is taken out randomly. What is the probability that the ball is neither white nor green?

Correct Answer: (a) 1/9  
Solution:

Total number of balls = (5 + 7 + 3) = 15
Number of red balls = 5
∴ Probability that the ball is neither white nor green
= 5/15 = 1/3

5. A container contains 26 yellow, 6 red, 13 green, and 19 pink marbles. 20 marbles are added to the container randomly. How many of the added marbles are yellow if the chance of choosing a yellow marble out of the container becomes 0.5.

Correct Answer: (e) 16
Solution:Total marbles = 16 + 6 + 23 + 19 = 64
After adding 20 marbles total marbles = 84
Let x be yellow marbles added to the container
As per the question
(x + 26)/84 = 0.5
⇒ x = 16
∴ 16 yellow marbles should be added

6. An urn contains 7 black, 5 red, 8 pink, and 4 orange marbles, if two marbles are picked at random, what is the probability that either both are red or orange?

Correct Answer: (d) 4/69  
Solution:Way to select two red marbles = ⁵C₂ = 10
Way to select 2 orange balls = ⁴C₂ = 6
∴ Required (P) = 10/²⁴C₂ + 6/²⁴C₂
⇒ 16/276 = 4/69
∴ The probability is 4/69

7. In how many ways 4 face cards can be selected from a well shuffled pack of 52 cards, such that at least 3 of them are of the same face?

Correct Answer: (e) None of these
Solution:

There are 4 face cards of jack, 4 face cards of queen and 4 face cards of king in a well shuffled pack of 52 cards.
So, number of ways of selecting 4 face cards such that at least 3 of them are of same face:
⇒ 3 × [⁴C₄ + (⁴C₃ × ⁴C₁) + (⁴C₃ × ⁴C₁)]
⇒ 3 × [1 + (4 × 4) + (4 × 4)] = 99
∴ The correct answer is 99.

8. A bag contains ‘x’ red and ‘x - 3’ green balls. If two balls are drawn at random from the bag one after another and without replacement, then the probability that both the balls are green in colour is (1/7). Find the probability of drawing a red ball from the bag.

Correct Answer: (e) None of these
Solution:Total number of balls in the bag = x + x − 3 = (2x − 3)
ATQ:
(x − 3)C₂ / (2x − 3)C₂ = (1/7)
[(x − 3)(x − 4)] / [(2x − 3)(2x − 4)] = (1/7)
7(x² − 7x + 12) = 4x² − 14x + 12
7x² − 49x + 84 = 4x² − 14x + 12
3x² − 35x + 72 = 0
3x² − 8x − 27x + 72 = 0
x(3x − 8) − 9(3x − 8) = 0
(x − 9)(3x − 8) = 0
x = 9 or x = (8/3) [not possible]
So, number of red balls in the bag = 9
Total number of balls in the bag = 2 × 9 − 3 = 15
Required probability = (9/15) = 0.6

9. The probability that it will not rain on Monday is (5/8) and the probability that it will rain on both Monday and Tuesday is (1/20). Find the probability that it will not rain on Tuesday.

Correct Answer: (c) (13/15)
Solution:Since, probability of occurrence of an event + probability of non-occurrence of an event = 1
Probability that it will rain on Monday = 1 − (5/8)
= (3/8)
Also, probability that it will rain on Monday × probability that it will rain on Tuesday = (1/20)
And so, probability that it will rain on Tuesday
= (1/20) ÷ (3/8) = (2/15)
And so, probability that it will not rain on Tuesday
= 1 − (2/15) = (13/15)

10. There are 30 balls in a bag. Number of red balls in the bag is same as number of yellow balls in the bag. If a ball is picked at random, then the probability that it’s a blue ball is 0.4. Find the number of red balls in the bag given that each ball in the bag is either red, blue or yellow.

Correct Answer: (a) 9  
Solution:Let number of blue balls in the bag be ‘x’
Therefore,
(x/30) = 0.4
Number of blue balls in the bag = ‘x’ = 30 × 0.4 = 12
Since, number of red balls = number of yellow balls
So, number of red balls in the bag = (30 − 12)/2 = 9