A die is thrown three times, and sum of three numbers thrown is 15. Find the chance that the first throw was a four.

\frac{1}{2} .

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\frac{1}{5} .

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\frac{1}{15} .

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\frac{1}{13} .

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Solution

The correct option is B $\frac{1}{5} .$
If first throw is four then the sum of numbers appearing on last two throws must be equal to $11$ .
That means last two throws are $(6, 5)$ or $(5, 6)$
Now, there are $10$ ways to get the sum as $15$ .i.e., ${(4, 5, 6), (4, 6, 5), . . ., (3, 6, 6)}$
$\Rightarrow$ Required probability $= \frac{2}{10} = \frac{1}{5}$

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A die is thrown three times, and sum of three numbers thrown is 15. Find the chance that the first throw was a four.

The correct option is B 15.If first throw is four then the sum of numbers appearing on last two throws must be equal to 11.That means last two throws are (6,5) or (5,6)Now, there are 10 ways to get the sum as 15.i.e., {(4,5,6),(4,6,5),...,(3,6,6)}⇒ Required probability =210=15