A pair of a dice is thrown, if $5$ appears on at least one of the dice, then the probability that the sum is $10$ or greater, is

\frac{11}{36}

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\frac{2}{9}

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\frac{3}{11}

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\frac{1}{12}

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Solution

The correct option is D $\frac{1}{12}$
If $5$ appears on at least one of the dice, then number of favourable cases of getting $10$ or greater than $10$ is $3$ ie, ${(5, 6) (5, 5) (6, 5)}$
Total number of cases $= 36$
$∴$ Required probability $= \frac{3}{36} = \frac{1}{12}$

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Similar questions

Two dice are thrown. What is the probability that the sum of the numbers appearing on the two dice is 11, if 5 appears on the first

(i)

(i i)

(i i i)

11

Event: 'Sum on $2$ dice'	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
Probability	$\frac{1}{36}$						$\frac{5}{36}$				$\frac{1}{36}$

11

2, 3, 4, 5, 6, 7, 8, 9, 10, 11

12

\frac{1}{11}

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