In a single throw of two dice, find the probability of obtaining
- a sum of less than 5;
- a sum of greater than 10;
- a sum of 9 or 11.
When two dice are thrown, number of elements in sample space $S = n\left(S\right) = 36$
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Let $A$ be the event of getting a sum of less than $5$
Then, $A = \left\{\left(1,1\right), \left(1,2\right), \left(1,3\right), \left(2,1\right), \left(2,2\right), \left(3,1\right)\right\}$
$\therefore \;$ Number of elements in $A = n\left(A\right) = 6$
$\therefore \;$ Probability of event $A = P\left(A\right) = \dfrac{n\left(A\right)}{n \left(S\right)} = \dfrac{6}{36} = \dfrac{1}{6}$
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Let $B$ be the event of getting a sum greater than $10$
Then, $B = \left\{\left(5,6\right), \left(6,5\right), \left(6,6\right)\right\}$
$\therefore \;$ Number of elements in $B = n\left(B\right) = 3$
$\therefore \;$ Probability of event $B = P\left(B\right) = \dfrac{n\left(B\right)}{n \left(S\right)} = \dfrac{3}{36} = \dfrac{1}{12}$
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Let $C$ be the event of getting a sum of $9$ or $11$
Then, $C = \left\{\left(3,6\right), \left(4,5\right), \left(5,4\right), \left(5,6\right), \left(6,3\right), \left(6,5\right)\right\}$
$\therefore \;$ Number of elements in $C = n\left(C\right) = 6$
$\therefore \;$ Probability of event $C = P\left(C\right) = \dfrac{n\left(C\right)}{n \left(S\right)} = \dfrac{6}{36} = \dfrac{1}{6}$