IIT JEE 1982 Maths - MCQ Question 8

Algebra Level 3

If A and B are events such that P(A)>0 and P(B)≠1 then P(A'/B') is equal to (where S' is the complement of S) which of the option?

In case you are preparing for IIT JEE, you may want to try IIT JEE 1982 Mathematics Archives

1 − P ( A ∪ B ) P ( B ′ ) \frac{1-P(A\cup B)}{P(B')} 1-P(A'/B) P(A')/P(B) 1-P(A/B)

Shubhamkar Ayare by Shubhamkar Ayare , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Tom Engelsman
Dec 23, 2016

Assuming A and B are independent events:

P ( A ′ ∣ B ′ ) = P ( A ′ ∩ B ′ ) P ( B ′ ) = P ( A ′ ) ⋅ P ( B ′ ) P ( B ′ ) = [ 1 − P ( A ) ] [ 1 − P ( B ) ] P ( B ′ ) = 1 − P ( A ) − P ( B ) + P ( A ∩ B ) P ( B ′ ) = 1 − [ P ( A ) + P ( B ) − P ( A ∩ B ) ] P ( B ′ ) = 1 − P ( A ∪ B ) P ( B ′ ) . P(A'|B') = \frac{P(A' \cap B')}{P(B')} = \frac{P(A') \cdot P(B')}{P(B')} = \frac{[1-P(A)][1-P(B)]}{P(B')} = \frac{1 - P(A) - P(B) + P(A \cap B)}{P(B')} = \frac{1 - [P(A) + P(B) - P(A \cap B)]}{P(B')} = \boxed{\frac{1 - P(A \cup B)}{P(B')}}.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...