GATE IT 2008 | Question: 2

Question

A sample space has two events $A$ and $B$ such that probabilities $P(A\cap B) = \dfrac{1}{2}, P(A') = \dfrac{1}{3}, P(B') =\dfrac{1}{3}$. What is $P(A\cup B)$ ?

• A. $\left(\dfrac{11}{12}\right)$
• B. $\left(\dfrac{10}{12}\right)$
• C. $\left(\dfrac{9}{12}\right)$
• D. $\left(\dfrac{8}{12}\right)$

Answer 1

Answer is (B) 

$P(A\cup B) = P(A) + P(B) - P(A\cap B)$

$= (1 - P(A')) + (1- P(B')) - P(A\cap B)$

$= \left(1-\dfrac{1}{3}\right) + \left(1-\dfrac{1}{3}\right) - \dfrac{1}{2}$

$=\dfrac{4}{3} - \dfrac{1}{2} =\dfrac{5}{6} =\dfrac{10}{12}.$

Here: If we simplify the option(B) we get $5/6.$

Comments

@omesh-pandita fine?

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yup

Answer 2

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