Fill in the blanks in following table: P(A) P(B) P(A ∩ B) P(A ∪ B) (i) … (ii) 0.35 … 0.25 0.6 (iii) 0.5 0.35 … 0.7

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Solution

(i)

The given probabilities are,

P( A )= 1 3 P( B )= 1 5 P( A∩B )= 1 15

We know that,

P( A∪B )=P( A )+P( B )−P( A∩B )

Substitute values of P( A ), P( B ) and P( A∩B ) in above formula.

P( A∪B )= 1 3 + 1 5 − 1 15 = 5+3−1 15 = 7 15

Thus, the value of P( A∪B ) is 1 15 .

(ii)

The given probabilities are,

P( A )=0.35 P( A∩B )=0.25 P( A∪B )=0.6

We know that,

P( A∪B )=P( A )+P( B )−P( A∩B )

Substitute values of P( A ), P( A∪B ) and P( A∩B ) in above formula.

0.6=0.35+P( B )+0.25 P( B )=0.6+0.25−0.35 =0.5

Thus, the value of P( B ) is 0.5.

(iii)

The given probabilities are,

P( A )=0.5 P( B )=0.35 P( A∪B )=0.7

We know that,

P( A∪B )=P( A )+P( B )−P( A∩B )

Substitute values of P( A ), P( B ) and P( A∪B ) in above formula.

0.7=0.5+0.35 P( A∩B )=0.5+0.35−0.7 =0.15

Thus, the value of P( A∩B ) is 0.15.

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Q. (a) If A and B be mutually exclusive events associated with a random experiment such that P(A) = 0.4 and P(B) = 0.5, then find:
(i) P (A ∪ B)
(ii)

P (A \cap B)

(iii) P (

A

∩ B)
(iv) P (A ∩

B

).

(b) A and B are two events such that P (A) = 0.54, P (B) = 0.69 and P (A ∩ B) = 0.35.
Find (i) P (A ∪ B)
(ii)

P (A \cap B)

(iii) P (A ∩

B

)
(iv) P (B ∩

A

)

(c) Fill in the blanks in the following table:

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(i)	$\frac{1}{3}$	$\frac{1}{5}$	$\frac{1}{15}$	......
(ii)	0.35	....	0.25	0.6
(iii)	0.5	0.35	.....	0.7

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