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In a Simultaneous Throw of a Pair of Dice, Find the Probability of Getting: a Sum More than 7 - Mathematics

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Question

In a simultaneous throw of a  pair of dice, find the probability of getting: 

a sum more than 7

Solution

\[\text{ When a pair of dice is thrown simultaneously, the sample space will be as follow  }: \]
\[S = \left\{ \left( 1, 1 \right), \left( 1, 2 \right), \left( 1, 3 \right), \left( 1, 4 \right), \cdots\left( 6, 5 \right), \left( 6, 6 \right) \right\}\]
\[\text{ Hence, the total number of outcomes is 36 } . \]

\[\text{ Let A be the event of getting pairs whose sum is more than 7 }. \]
\[\text{ The pairs whose sum is more than 7 are  } \left( 2, 6 \right), \left( 3, 5 \right), \left( 3, 6 \right), \left( 4, 4 \right), \left( 4, 5 \right), \left( 4, 6 \right), \left( 5, 3 \right), \left( 5, 4 \right), \left( 5, 5 \right), \left( 5, 6 \right), \left( 6, 2 \right), \left( 6, 3 \right), \left( 6, 4 \right), \left( 6, 5 \right) \text{ and  }\left( 6, 6 \right) . \]
\[\text{ Hence, the number of favourable outcomes is }15 . \]
\[ \therefore P\left( A \right) = \frac{\text{ Number of favourable outcomes }}{\text{ Total number of outcomes }} = \frac{15}{36} = \frac{5}{12}\]

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Chapter 26: Data Handling-IV (Probability) - Exercise 26.1 [Page 15]

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RD Sharma Mathematics [English] Class 8
Chapter 26 Data Handling-IV (Probability)
Exercise 26.1 | Q 3.11 | Page 15

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