Question

Question 7 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′∩B′) (iii) P(A∩B′) (iv) P(B∩A′)

Class X1 - Maths -Probability Page 409

Answer 1

Using the relations :
( a ) P( A U B ) = P( A ) + P( B ) - P( A ∩ B )
( b ) P( A' ∩ B') = P(A U B)' = 1 - P( A U B )
( c ) P( A ∩ B') = P( A ) - P( A ∩ B)
( d ) P( B ∩ A') = P( B ) - P( B ∩ A)

Here, P( A ) = 0.54 , P( B ) = 0.69
P( A ∩ B ) = 0.35

(i) P( A U B ) = P( A ) + P( B ) - P( A ∩ B )
= 0.54 + 0.69 - 0.35
= 1.23 - 0.35
= 0.88

( ii) P( A' ∩ B') = P(A U B)' = 1 - P( A U B )
= 1 - 0.88
= 0.12

(iii) P( A ∩ B') = P( A ) - P( A ∩ B)
= 0.54 - 0.35
= 0.19

( iv) P( B ∩ A') = P( B ) - P( B ∩ A)
= 0.69 - 0.35
= 0.34

Answer 2

It is given that P(A) = 0.54, P(B) = 0.69, P(A n B) = 0.35

(i) We know that P (A D B) = P(A) + P(B) - P(A n B)

P (A D B) = 0. 54 + 0.69 - 0.35 = 0.88

(ii ) A' n B' = (A DB)' [by De Morgan's law]

P(A' n B') = P(A D B)' = 1 - P(A D B) = 1 - 0.88 = 0.12

(iii) P(A n B') = P(A) - P(A n B) = 0.54 - 0.35 = 0.19

(iv) We know tha n(BnA') = n(B)-n(AnB)

n(BnA')/n (S) = n(b)/n (S) - n(AnB)/n (S)

P(BnA') = P(B)-P(AnB)

P(BOA') =0.69-0.35 =0.34

I HOPE IT WILL HELP YOU.

THANK YOU.