#### Prev Next

**Probability:**

- Event: Any subset of a sample space is called an event.
- Probability of Occurrence of an Event: Let S be the sample space and let E be an event.

Then, E⊆S

∴ P(E) = n(E)′/n (S) - Sample space: When we perform an experiment, then the set S of all possible outcomes is called the Sample space.

**Aptitude Questions (**Probability**):**

1. In a simultaneous throw of two coins, the probability of getting at least one head is:

A) (1/2)

B) (1/3)

C) (2/3)

D) (3/4)

E) None of these

View Answer

Answer: option (D)

Explanation:

Here S = {HH, HT, TH, TT}

Let E = event of getting at le* one head = {HT, TH, HH}

∴ P(E) = (n(E)/n(S)) = (3/4)

2. Three unbiased coins are tossed. What is the probability of getting at least 2 heads?

A) (1/4)

B) (1/2)

C) (1/3)

D) (1/8)

E) None of these

View Answer

Answer: option (B)

Explanation:

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}

∴ P(E) = (n(E)/n(S)) = (4/8) = (1/2)

3. Three unbiased coins are tossed. What is the probability of getting at most 2 heads?

A) (3/4)

B) (1/4)

C) (3/8)

D) (7/8)

E) None of these

View Answer

Answer: option (D)

Explanation:

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at most two heads

Then, E = {TTT, TTH, THT, HTT, THH, HTH, HHT}

∴ P(E) = (n(E)/n(S)) = (7/8)

4. In a single throw of a dice, what is the probability of getting a number greater than 4?

A) (1/2)

B) (1/3)

C) (2/3)

D) (1/4)

E) None of these

View Answer

Answer: option (B)

Explanation:

When a dice is thrown, we have S = {1, 2, 3, 4, 5, 6}

Let E = event of getting a number greater than 4 = {5, 6}

∴ P(E) = (n(E)/n(S)) = (2/6) = (1/3)

5. In a simultaneous throw of two dice, what is the probability of getting a total of 7?

A) (1/6)

B) (1/4)

C) (2/3)

D) (3/4)

E) None of these

View Answer

Answer: option (A)

Explanation:

We know that in a simultaneous throw of two dice, n(S) = 6 * 6 = 36

Let E = event of getting a total of 7 = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

∴ P(E) = (n(E)/n(S)) = (6/36) = (1/6)