Probability: Bank Exam Practice Questions (SBI, IBPS, RRB, PO & Clerk)

Ans: C

Explanation: Let ‘g’ be the number of green balls and ‘b’ be the number of blue balls. Let ‘t’ be the total number of balls. We know that there are 17 red balls.
The probability of picking a green ball is g/t. The probability of picking a red ball is 17/t.
According to the question, g/t = 17/t + 1/7
The probability of picking a blue ball is b/t.
b/t = 17/t + 1/21
We also know that t = 17 + g + b.
From the first equation, multiply by 7t: 7g = 119 + t, or t = 7g – 119
From the second equation, multiply by 21t: 21b = 357 + t, or t = 21b – 357
Since t = 17 + g + b, substitute t for 7g – 119, so 7g – 119 = 17 + g + b
6g – b = 136. Also, since t = 21b – 357, then 17 + g + b = 21b – 357.
g – 20b = -374.
From 6g – b = 136, we can write g = (136 + b)/6. Substitute g in the previous equation.
(136 + b)/6 – 20b = -374
136 + b – 120b = -2244
-119b = -2380
b = 20.
Let’s find g.
g = (136 + 20)/6 = 156/6 = 26
t = 17 + 26 + 20 = 63.
P(red) = 17/63
P(green) = 26/63 = 17/63 + 1/7 = 17/63 + 9/63 = 26/63
P(blue) = 20/63 = 17/63 + 1/21 = 17/63 + 3/63 = 20/63
Therefore the number of blue balls is 20.

Correct Option: C

Ans: C

Explanation:
First, calculate the probability of drawing a red ball from the first bag: 4 red balls / (4 red + 6 blue) = 4/10 = 2/5.
Second, calculate the probability of drawing a red ball from the second bag: 3 red balls / (3 red + 7 blue) = 3/10.
Third, assume that the probability of choosing each bag is 1/2.
So, the overall probability of drawing a red ball is (1/2 * 2/5) + (1/2 * 3/10) = 2/10 + 3/20 = 4/20 + 3/20 = 7/20.

Ans: B

Explanation: There are two possible scenarios to get one pink and one blue ball: picking a pink ball first then a blue ball, or picking a blue ball first then a pink ball.
Scenario 1: Pink then Blue
Probability of picking a pink ball first: 7/16
Probability of picking a blue ball second (given a pink was picked first): 9/15
Probability of Scenario 1: (7/16) * (9/15) = 63/240
Scenario 2: Blue then Pink
Probability of picking a blue ball first: 9/16
Probability of picking a pink ball second (given a blue was picked first): 7/15
Probability of Scenario 2: (9/16) * (7/15) = 63/240
Total Probability = Probability of Scenario 1 + Probability of Scenario 2 = 63/240 + 63/240 = 126/240
Simplify the fraction: 126/240 = 21/40

Correct Option: B

Ans: D

Explanation: Let X be the number of burnt muffins in a box of 4. We want to find the probability P(X <= 1). This is equal to P(X=0) + P(X=1).
The probability of a muffin being burnt is 0.2, so the probability of a muffin not being burnt is 1 – 0.2 = 0.8.

P(X=0): All 4 muffins are not burnt. This is (0.8)^4 = 0.4096
P(X=1): Exactly one muffin is burnt, and the other 3 are not. There are 4 ways this can happen (the burnt muffin can be in any of the 4 positions). This probability is 4 * (0.2) * (0.8)^3 = 4 * 0.2 * 0.512 = 0.4096
P(X <= 1) = P(X=0) + P(X=1) = 0.4096 + 0.4096= 0.8192

Ans: A

Explanation:
1. **Identify the original face cards:** A standard deck has 12 face cards (Jack, Queen, King of each suit).
2. **Remove the red face cards:** The red face cards are the Jack, Queen, and King of Hearts and Diamonds. That’s 6 cards.
3. **Face cards remaining:** After removing the red face cards, there are 6 face cards left (Jack, Queen, King of clubs and spades).
4. **Cards remaining in the deck:** The original deck had 52 cards. 6 cards were removed (the red face cards). The deck now has 52 – 6 = 46 cards.
5. **Calculate the probability:** The probability of choosing a face card is the number of remaining face cards divided by the total number of cards remaining: 6/46 = 3/23

Correct Option: A

Ans: E

Explanation: A standard die has 6 sides, numbered 1 through 6. The even numbers are 2, 4, and 6. There are 3 even numbers out of 6 possible outcomes. Therefore, the probability of rolling an even number is 3/6, which simplifies to 1/2.

Ans: E

Explanation:
Quantity A: To form a team with exactly 3 batsmen, we must also select 1 bowler. The number of ways to choose 3 batsmen from 6 is 6C3 = (6!)/(3!3!) = 20. The number of ways to choose 1 bowler from 4 is 4C1 = 4. So the number of ways to form the team with exactly 3 batsmen is 20 * 4 = 80.

Quantity B: To form a team with at least 3 bowlers, we can have either 3 bowlers and 1 batsman or 4 bowlers and 0 batsmen.
– 3 bowlers and 1 batsman: 4C3 * 6C1 = 4 * 6 = 24
– 4 bowlers and 0 batsmen: 4C4 * 6C0 = 1 * 1 = 1
So the number of ways to form the team with at least 3 bowlers is 24 + 1 = 25.

Comparing Quantity A (80) and Quantity B (25), Quantity A > Quantity B.

Ans: A

Explanation: The probability of getting tails on a single coin flip is 1/2. The probability of getting three tails in a row is (1/2) * (1/2) * (1/2) = 1/8. Comparing this to 1/16, we have 1/8 > 1/16. Therefore, Quantity I is greater than Quantity II.
Correct Option: A

Ans: C

Explanation: There are 12 face cards in a standard deck of cards (Jack, Queen, King for each of the four suits). The red suits are hearts and diamonds, and the black suits are clubs and spades. This means there are 6 red face cards and 6 black face cards. We need to find the number of combinations with at least two red face cards. This breaks down into two cases:

Case 1: Exactly two red face cards.
We choose 2 red face cards from the 6 red face cards, and 1 black face card from the 6 black face cards. The number of combinations is (6 choose 2) * (6 choose 1) = (6!/(2!4!)) * 6 = (15) * 6 = 90

Case 2: Exactly three red face cards.
We choose 3 red face cards from the 6 red face cards. The number of combinations is (6 choose 3) = 6!/(3!3!) = (6*5*4)/(3*2*1) = 20

Total number of combinations with at least two red face cards is 90 + 20 = 110.

Correct Option: C

Ans: A

Explanation: First, determine all the possible outcomes when rolling two dice. There are 6 sides on each die, so there are 6 * 6 = 36 total possible outcomes.
Next, find the outcomes where the sum of the numbers rolled is greater than 9. These are:
* (4, 6)
* (5, 5)
* (5, 6)
* (6, 4)
* (6, 5)
* (6, 6)
There are 6 favorable outcomes.

The probability is the number of favorable outcomes divided by the total number of outcomes: 6/36 = 1/6.

Ans: E

Explanation:
Let’s break down the problem step by step.

**Step 1: Analyze Office P in March**

* Office P starts with 6 males and 4 females, totaling 10 employees.
* The probability of selecting a male candidate initially is 6/10 = 3/5.

**Step 2: Analyze the Increase in Probability in April**

* The probability of selecting a male candidate increases by 4 and 1/6%, which is 4.166…% or 1/24.
* Let ‘x’ be the number of new employees in Office P.
* The number of male employees increases by ‘m’ and the number of female employees increases by ‘f’. So m+f=x.
* The new total number of male employees is 6+m. The new total is 10+x.
* The new probability of selecting a male is (6+m)/(10+x) = 3/5 + 1/24 = 72/120 + 5/120 = 77/120.

We also know that (6+m)/(10+x)= 77/120. If we are only provided with the information of increase, we may assume no new female employees join Office P, where x=m. Then (6+x)/(10+x)=77/120.
Cross multiply, 120(6+x)=77(10+x)
720+120x=770+77x
43x = 50
x=50/43. As x must be an integer, it is reasonable to think that the increase is simply due to the change in the total number of people in Office P.

However, assuming the total number of employees in office P is 50/43 is also not reasonable. We should make the probability of picking a male, (6+m)/(10+m+f), increase by 1/24. Assuming all new employees are male, then (6+x)/(10+x)=77/120, x=50/43. This is not possible.
Let us look at some standard ratios:
(6+m)/(10+x) = 77/120, let’s suppose m=0 and x=0, (6/10)<77/120. We will assume some new male employees have joined. We assume the new probability is achieved by adding only male employees. We assume (6+a)/(10+a)=77/120.
So, 120*6 + 120a=77*10+77a
720+120a=770+77a
43a=50, which doesn’t give integer values.
We are missing some information here.
If the number of new people is 5, and all 5 are male, then P has 11 males and 4 females. 11/15 is not close to 77/120.
If the number of new people is 5, 2 males, 3 females, the new male probability is (6+2)/(10+5)=8/15, which is not close to 77/120.

**Step 3: Analyzing the Transfers**

* A person is moved from Office P to Q, and then a person is moved from Office Q to P. We must consider all cases.

Let’s assume the question meant that only a single person moves.
**Scenario 1: A male is moved from P to Q, and a male is moved from Q to P**
Initially P: 6 males, 4 females. Q: 4 males, 6 females.
P: 5 males, 4 females. Q: 5 males, 6 females.
P: 6 males, 4 females. Q: 4 males, 6 females. The probability of picking male candidate is 6/10 = 3/5.

**Scenario 2: A male is moved from P to Q, and a female is moved from Q to P**
Initially P: 6 males, 4 females. Q: 4 males, 6 females.
P: 5 males, 4 females. Q: 5 males, 6 females.
P: 5 males, 5 females. Q: 4 males, 5 females. The probability of picking male candidate is 5/10=1/2.

**Scenario 3: A female is moved from P to Q, and a male is moved from Q to P**
Initially P: 6 males, 4 females. Q: 4 males, 6 females.
P: 6 males, 3 females. Q: 4 males, 7 females.
P: 7 males, 3 females. Q: 3 males, 7 females. The probability of picking male candidate is 7/10.

**Scenario 4: A female is moved from P to Q, and a female is moved from Q to P**
Initially P: 6 males, 4 females. Q: 4 males, 6 females.
P: 6 males, 3 females. Q: 4 males, 7 females.
P: 6 males, 4 females. Q: 4 males, 6 females. The probability of picking male candidate is 6/10 = 3/5.
Since there is not additional data about the exact number of new employees, and the final state is not specified, it is impossible to determine the final probability precisely based on the available information. Assuming a single person is moved.

If we suppose m=0 and x=5, then P would contain 6M and 4F. P would gain 5, 6/10 +5/10*0 + 5/10*0
So the final probability would depend on x and the transfer.

**Revised Attempt**
Let’s try to proceed by assuming single person transfers. Initially, P has 6M/10. The increase in probability is 1/24. 6/10+1/24 = 77/120. This indicates the question intended for a new scenario. It is still hard to determine.

Not able to solve.

Explanation: The question is missing necessary information, such as the actual composition and number of new employees added to office P in April. The transfer details also require the exact number of individuals moved.

Correct Option: Not able to solve

Ans: B

Explanation: Let J be the event that John completes the project successfully, and M be the event that Mary completes the project successfully. We are given P(J) = 0.6 and P(M) = 0.5. Since their efforts are independent, the probability that both fail is P(J’) * P(M’) = (1-0.6) * (1-0.5) = 0.4 * 0.5 = 0.2. The probability that at least one of them completes the project successfully is the complement of the probability that both fail, which is 1 – 0.2 = 0.8.

Ans: C

Explanation: The probability that A is not selected is 1 – (1/5) = 4/5. The probability that B is not selected is 1 – (1/6) = 5/6. Assuming A and B’s selection are independent events, the probability that neither is selected is (4/5) * (5/6) = 20/30 = 2/3.

Correct Option: C