A die is rolled three times. The probability that the sum of three numbers obtained is $15,$ is equal to :

\frac{5}{108}

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\frac{5}{216}

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\frac{11}{216}

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\frac{7}{108}

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\frac{13}{216}

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Solution

The correct option is A $\frac{5}{108}$
Total number of cases $= 6^{3} = 216$

The sum of $15$ is obtained by three dice in following ways :
(i) $(3, 6, 6) = 3$ ways
(ii) $(4, 5, 6) = 6$ ways
(iii) $(5, 5, 5) = 1$ way

Therefore, favourable number of ways $= 10$

Thus required probability is $\frac{10}{216} = \frac{5}{108}$ .

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A die is rolled three times. The probability that the sum of three numbers obtained is 15, is equal to :

The correct option is A 5108Total number of cases =63=216The sum of 15 is obtained by three dice in following ways :(i) (3,6,6)=3 ways(ii) (4,5,6)=6 ways(iii) (5,5,5)=1 wayTherefore, favourable number of ways =10Thus required probability is 10216=5108.

A die is rolled three times. The probability that the sum of three numbers obtained is $15,$ is equal to :