Probability

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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)

 

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