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