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Question

Given two independent events A and B such that P (A) = 0.3, P (B) = 0.6. Find (i) P (A and B) (ii) P (A and not B) (iii) P (A or B) (iv) P (neither A nor B)

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Solution

(i)

The probability of A is 03 and the probability of B is 06.

The formula for two independent events is,

P( AB )=P( A )P( B )

Substitute the given values in the above formula we get,

P( AB )=03×06 =018

Thus, the value of P( AB )=018.

(ii)

The probability of ( A and not B ) or P( AB' ) is,

P( A B )=P( A )P( AB ) =0.30.18 =0.12

Thus, the value of P( AB' )=0.12.

(iii)

The formula of P( AB ) is,

P( AB )=P( A )+P( B )P( AB )

Substitute the given values in the above formula we get,

P( AB )=0.3+0.60.18 =0.90.18 =0.72

Thus, the value of P( AB )=0.72.

(iv)

The formula probability of (neither A nor B) or P( A B ) is,

P( A B )=P ( AB ) =1P( AB )

Substitute the given values in the above formula we get,

P( A B )=10.72 =0.28

Thus, the value of P( A B )=0.28.


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