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A-die-tossed-twice-find-probability-getting-number-greater-3-each-toss - Mathematics

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प्रश्न

A die is tossed twice. Find the probability of getting a number greater than 3 on each toss.

 
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उत्तर

\[S = { \left( 1 1 \right), \left( 1 2 \right), \left( 1 3 \right), \left( 1 4 \right), \left( 1 5 \right), \left( 1 6 \right),} \]  \[ \left( 2 1 \right), \left( 2 2 \right), \left( 2 3 \right), \left( 2 4 \right), \left( 2 5 \right), \left( 2 6 \right), \]
\[ \left( 3 1 \right), \left( 3 2 \right), \left( 3 3 \right), \left( 3 4 \right), \left( 3 5 \right), \left( 3 6 \right), \]
\[ \left( 4 1 \right), \left( 4 2 \right), \left( 4 3 \right), \left( 4 4 \right), \left( 4 5 \right), \left( 4 6 \right), \]
\[ \left( 5 1 \right), \left( 5 2 \right), \left( 5 3 \right), \left( 5 4 \right), \left( 5 5 \right), \left( 5 6 \right), \]
\[ \left( 6 1 \right), \left( 6 2 \right), \left( 6 3 \right), \left( 6 4 \right), \left( 6 5 \right), \left( 6 6 \right) \]
\[n\left( S \right) = 36\]
\[E = \text{ Getting a number greater than 3 on each toss } \]
\[ = \left\{ \left( 4 4 \right), \left( 4 5 \right), \left( 4 6 \right), \left( 5 4 \right), \left( 5 5 \right), \left( 5 6 \right), \left( 6 4 \right), \left( 6 5 \right), \left( 6 6 \right) \right\}\]
\[n\left( E \right) = 9\]
\[P\left( E \right) = \frac{9}{36}\]
\[ = \frac{1}{4}\]

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 31: Probability - Exercise 31.4 [पृष्ठ ५४]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 31 Probability
Exercise 31.4 | Q 11 | पृष्ठ ५४
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