Let A and B be two independent events such
that
P(A) = ${1 \over 3}$ and P(B) = ${1 \over 6}$.
Then, which of
the following is TRUE?
Solution
Given P(A) = ${1 \over 3}$ and P(B) = ${1 \over 6}$
<br><br>A and B are independent
<br><br>So P(A$\cap$B) = ${1 \over 3} \times {1 \over 6}$ = ${1 \over {18}}$
<br><br>P(A$\cup$B) = P(A) + P(B) – P(A $\cap$ B)
<br><br>= ${1 \over 3}$ + ${1 \over 6}$ - ${1 \over {18}}$
<br><br>= ${4 \over 9}$
<br><br>$P\left( {{A \over {B'}}} \right)$
<br><br>= ${{P\left( {A \cap B'} \right)} \over {P\left( {B'} \right)}}$
<br><br>= ${{P\left( A \right) - P\left( {A \cap B} \right)} \over {1 - P\left( B \right)}}$
<br><br>= ${{{1 \over 3} - {1 \over {18}}} \over {1 - {1 \over 6}}}$ = ${{1 \over 3}}$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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