Probability Examples 2

In this class, We discuss Probability Examples 2.

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Example 1:

Three students, A, B, and C, are in a running race.

A and B have the same probability of winning, and each is twice as likely to win as C.

Find the probability B or C wins.

Solution:

A U B U C is the sample space.

P(A) = P(B)

P(A) = 2P(C) = P(B)

P(A) + P(B) + P(C) = 1 because A U B U C is the sample space.

Substitute the values.

2P(C) + 2P(C) + P(C) = 1

5P(C) = 1

P(C) = 1/5

P(A) = P(B) = 2/5

Probability of B or C wins?

Both events B and C are mutually exclusive because either one will win, not both.

P(B U C) = P(B) + P(C) – P(B ∩ C)

P(B ∩ C) = 0 because mutually exclusive events.

P(B U C) = 2/5 + 1/5

P(B U C) = 3/5

Example 2:

From a city, three newspapers A, B, C.

20% of them read newspaper A.

16% of them read newspapers. B.

14% of them read the newspaper C.

Both A and B read by 8%.

Both A and C read by 5%.

Both B and C read by 4%.

Both A and B and C read by 2%.

What is the percentage of the population that read atleast one newspaper?

Solution:

P(A U B U C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C)

P(A) = 20/100

P(B) = 16/100

P(C) = 14/100

P(A ∩ B) = 8/100

P(A ∩ C) = 5/100

P(B ∩ C) = 4/100

P(A ∩ B ∩ C) = 2/100

Substitute in the equation.

P(A U B U C) = 35/100