Problem 1 :
If A and B are mutually exclusive events P(A) = 3/8 and P(B) = 1/8, then find (i) P(A') (ii) P(A U B) (iii) P(A' n B) (iv) P(A' U B')
Solution :
(i) P(A')
P(A') = 1 - P(A)
= 1 - (3/8)
P(A') = 5/8
(ii) P(A U B)
P(A U B) = P(A) + P(B) - P(A n B)
Since A and B are mutually exclusive events, P(A n B) = 0.
P(A U B) = (3/8) + (1/8)
= (3 + 1)/8
= 4/8
P(A U B) = 1/2
(iii) P(A' n B)
P(A' n B) = P(B) - P(A n B)
= (1/8) - 0
P(A' n B) = 1/8
(iv) P(A' U B')
P(A' U B') = P(A n B)'
P(A n B)' = 1 - P(A n B)
= 1 - 0
P(A n B)' = 1
Problem 2 :
If A and B are two events associated with a random experiment for which P(A) = 0.35, P(A or B) = 0.85, and P(A and B) = 0.15. Find (i) P(only B) (ii) P(B') (iii) P(only A)
Solution :
Given that :
P(A) = 0.35, P(A or B) = 0.85, and P(A and B) = 0.15.
(i) P(only B)
First let us find the value of P(B), for that let use the formula for addition theorem on probability.
P(AUB) = P(A) + P(B) - P(A n B)
0.85 = 0.35 + P(B) - 0.15
0.85 = 0.20 + P(B)
0.85 - 0.20 = P(B)
P(B) = 0.65
P(only B) = P(A' n B)
= P(B) - P(AnB)
= 0.65 - 0.15
P(only B) = 0.50
(ii) P(B')
P(B') = 1 - P(B)
= 1 - 0.65
P(B') = 0.35
(iii) P(only A)
P(only A) = P(AnB')
= P(A) - P(AnB)
= 0.35 - 0.15
P(only A) = 0.20
Problem 3 :
A die is thrown twice. Let A be the event, ‘First die shows 5’ and B be the event, ‘second die shows 5’. Find P(A∪B) .
Solution :
Sample space = 36
Let A be the event that the first die shows the number "5"
A = {(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)}
n(A) = 6
P(A) = n(A)/n(S)
P(A) = 6/36
Let B be the event that the second die shows the number "5"
B = {(1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5)}
n(B) = 6
P(B) = n(B)/n(S)
P(B) = 6/36
A n B = {5, 5}
n(AnB) = 1
P(AnB) = n(AnB)/n(S)
P(AnB) = 1/36
P(A U B) = P(A) + P(B) - P(A n B)
= (6/36) + (6/36) - (1/36)
= (12 - 1)/36
= 11/36
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