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Question
An unbiased die is tossed twice. Find the probability of getting 4, 5, or 6 on the first toss and 1, 2, 3 or 4 on the second toss.
Solution
\[P\left( A \right) = P\left( 4, 5 \text { or }6 \text{ on first toss } \right) = \frac{18}{36} = \frac{1}{2}\]
\[P\left( B \right) = P\left( 1, 2, 3 \text { or } 4 \text{ on second toss } \right) = \frac{24}{36} = \frac{2}{3}\]
\[\text{ It is clear that A and B are independent events.} \]
\[ \Rightarrow P\left( A \cap B \right) = P\left( A \right)P\left( B \right)\]
\[ \Rightarrow P\left( A \cap B \right) = \frac{1}{2} \times \frac{2}{3}\]
\[ \therefore P\left( A \cap B \right) = \frac{1}{3}\]
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