Probability refers to the quantitative measure of the likelihood of an event occurring. It is either expressed in percentages or proportions in the range 0% to 100% or 0 to 1 respectively.

Here are basic formulas and concepts that help compute the probability of an event:

Probability Range | 0 = P(A) = 1 |

Rule of Complementary Events | P(AC) + P(A) = 1 |

Rule of Addition | (A or B) = P(A) + P(B) – P(AnB) |

Disjoint Events | P(AnB) = 0 |

Conditional Probability | P(A | B) = P(AnB) / P(B) |

Bayes Formula | P(A | B) = P(B | A) × P(A) / P(B) |

Independent Events | P(AnB) = P(A) × P(B) |

Cumulative Distribution Function | FX(x) = P(X = x) |

We also use concepts of permutations and combinations to find the probability of an event.

**Checkout: **Guesstimate Interview Questions & Answers

Now, let’s look at some example probability aptitude questions and answers so you know how different types of questions in probability are approached.

**Probability Aptitude Questions and Answers**

**Question 1**: Two brothers X and Y appeared for an exam. Let A be the event that X is selected and B is the event that Y is selected.

The probability of A is 1/8 and that of B is 1/7. Find the probability that both of them are selected.

A) 1/63

B) 2/65

C) 1/56

D) 9/76

**Solution**:

**Option C (1/56) is the right answer. **

Given, A is the event that X is selected and B is the event that Y is selected.

P(A)=1/8, P(B)=1/7

If C be the event that both are selected, then

P(C)=P(A)×P(B) {A and B are independent events}

= (1/8)×(1/7)

= 1/56

**Question 2**: A person begins with 128 rupees and makes 6 bets, winning three times and losing three times, in random order. Both chances of winning and losing are equal. If each wager is for half the money remaining at the time of the bet, then the final result is:

A) gain of Rs 54

B) a loss of Rs 74

C) neither gain nor a loss

D) a gain or a loss depending upon the order in which the wins and losses occur

**Solution**:

**Option B is the right answer. **

Each win implies multiplying the amount by 1.5 and loss implies multiplying the amount by 0.5. Therefore, we will multiple 1.5 three times with the initial amount as well as multiply 0.5 three times.

The resultant amount in each case will be:

⇒ 128(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs 54

Hence the final result is:

⇒ 128 − 54 = 74

I.e., a loss of Rs 74

**Question 3:** A box contains 2 blue, 3 red, and 2 green balls. If a ball is drawn randomly, what is the probability that none of the balls is green?

A) 10/12

B) 10/21

C) 10/13

D) 10/31

**Solution**:

**Option B is the right answer.**

Number of balls in the box = (2+3+2) = 7

If S is the sample space.

Then, n(S) is the total number of ways you can draw 2 balls out of 7

⇒ 7C2 = (7×6)/(2×1) = 21

Let E be the event of drawing 2 balls out of the box, which isn’t green.

⇒ n(E) = number of ways you can draw 2 balls out of the remaining (2+3) balls in the box

= 5C2

⇒ (5×4)/2 = 10

Therefore, P(E) = n(E)/n(S) = 10/21

**Question 4:** If a bag contains 23 toys having numbers 1 to 23 and two toys are drawn at random one after another (without replacement of the first toy), what is the probability that both toys will have even numbers?

A) 5/21

B) 9/42

C) 11/42

D) 5/23

**Solution:**

**Option D is the right answer**.

There are 11 even numbers between 1 and 23. Therefore, the probability that the first toy shows have an even number is

= 11/23

Since the first toy is not replaced, there are 10 even number of toys left in the bag is a total of 22 toys.

Hence, the probability that the second toy is numbered even is,

= 10/22

Resultant probability =11/23)×(10/22)

=5/23

**Question 5**: X speaks the truth in 75% of cases and Y speaks the truth in 80% of cases. What is the percentage of cases that are likely to contradict each other while narrating the same event?

A) 45%

B) 5%

C) 35%

D) 22.5%

**Solution**:

**Option C is the right answer. **

There can be two cases of X and Y contradicting each other,

Case I: X speaks the truth and Y does not.

Case II: X does not speak the truth and Y speaks the truth.

⇒ (3/4×1/5) + (1/4×4/5) = 3/20 + 4/20 = 7/20

Therefore, the percentage is 35%.

**Question 6**: Given a set of numbers {1, 2, 3, …., 125}. If a number is chosen at random from the above set, what is the probability that the chosen number will be a perfect cube?

A) 1/2

B) 1/25

C) 4/13

D) 1/10

**Solution**:

**Option B is the right answer**.

There are five perfect cubes between 1 and 100 ⇒ 1, 8, 27, 64 and 125.

Therefore, the probability of picking a perfect cube from the given sample space is

= 5/125

= 1/25

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**Conclusion**

Acing a probability aptitude test requires practice and familiarity with the kind of questions commonly asked. Go through the basic probability formulas and concepts and practice as much as you can to feel confident about solving probability problems easily. Good luck!

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