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Question
A die is tossed twice. Find the probability of getting a number greater than 3 on each toss.
Solution
\[S = { \left( 1 1 \right), \left( 1 2 \right), \left( 1 3 \right), \left( 1 4 \right), \left( 1 5 \right), \left( 1 6 \right),} \] \[ \left( 2 1 \right), \left( 2 2 \right), \left( 2 3 \right), \left( 2 4 \right), \left( 2 5 \right), \left( 2 6 \right), \]
\[ \left( 3 1 \right), \left( 3 2 \right), \left( 3 3 \right), \left( 3 4 \right), \left( 3 5 \right), \left( 3 6 \right), \]
\[ \left( 4 1 \right), \left( 4 2 \right), \left( 4 3 \right), \left( 4 4 \right), \left( 4 5 \right), \left( 4 6 \right), \]
\[ \left( 5 1 \right), \left( 5 2 \right), \left( 5 3 \right), \left( 5 4 \right), \left( 5 5 \right), \left( 5 6 \right), \]
\[ \left( 6 1 \right), \left( 6 2 \right), \left( 6 3 \right), \left( 6 4 \right), \left( 6 5 \right), \left( 6 6 \right) \]
\[n\left( S \right) = 36\]
\[E = \text{ Getting a number greater than 3 on each toss } \]
\[ = \left\{ \left( 4 4 \right), \left( 4 5 \right), \left( 4 6 \right), \left( 5 4 \right), \left( 5 5 \right), \left( 5 6 \right), \left( 6 4 \right), \left( 6 5 \right), \left( 6 6 \right) \right\}\]
\[n\left( E \right) = 9\]
\[P\left( E \right) = \frac{9}{36}\]
\[ = \frac{1}{4}\]
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