# PRACTICE PROBLEMS ON ADDITION THEOREM OF PROBABILITY

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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