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A box has four dice in it. Three of them are fair dice but the fourth one has the number five on all of its faces. A die is chosen at random from the box and is rolled three times and shows up the face five on all the three occassions. The chance that the die chosen was a rigged die, is

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$\frac{216}{217}$

$\frac{215}{219}$

$\frac{216}{219}$

None of these

answer is C.

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Detailed Solution

Let $E_{1} :$ The event that fair die is chosen ..

$E_{2} :$ The event that rigged die is chosen

A : The event that tossing die three times it shows 5 on all the faces.

$\begin{aligned} P (E_{1}) = \frac{3}{4}, P (E_{2}) = \frac{1}{4} \\ P (\frac{A}{E_{1}}) =^{3} C_{3} {(\frac{1}{6})}^{3} = \frac{1}{6^{3}} \\ P (\frac{A}{E_{2}}) = 1 \end{aligned}$

Now, $P (\frac{E_{2}}{A}) = \frac{P (E_{2}) P (\frac{A}{E_{2}})}{P (E_{1}) P (\frac{A}{E_{1}}) + P (E_{2}) P (\frac{A}{E_{2}})}$

$= \frac{\frac{1}{4} \times 1}{\frac{3}{4} \times \frac{1}{6^{3}} + \frac{1}{4} \times 1} = \frac{1}{\frac{3}{216} + 1} = \frac{216}{219}$

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