Four fair dice $D_{1}, D_{2}, D_{3} a n d D_{4}$ each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously. The probability that $D_{4}$ shows a number appearing on one of $D_{1}, D_{2} a n d D_{3}$ , is ?

$\frac{91}{216}$

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$\frac{108}{216}$

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$\frac{125}{216}$

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$\frac{127}{216}$

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Solution

The correct option is A
$\frac{91}{216}$

Sample space $6 \times 6 \times 6 \times 6 = 6^{4}$ favourable events
= Case I or Case II or Case III
Case I First we should select one number for $D_{4}$ which apperars on all i.e. $^{6} C_{1} \times 1.$
Case II For $D_{4}$ there are $^{6} C_{1}$ ways. Now, it appears on any one of $D_{1}, D_{2}$ and $D_{3}$ i.e. $^{3} C_{1} \times 1$ .
For other two there are $5 \times 5$ ways.
$\Rightarrow^{6} C_{1} \times^{3} C_{1} \times 1 \times 5 \times 5$
Case III For $D_{4}$ there are $^{6} C_{1}$ ways now it appears on any two of $D_{1}, D_{2} a n d D_{3}$
$\Rightarrow^{3} C_{2} \times 1^{2}$
For other one there are 5 ways.
$\Rightarrow^{6} C_{1} \times^{3} C_{2} \times 1^{2} \times 5$
Thus, probability $= \frac{^{6} C_{1} +^{6} C_{1} \times^{3} C_{1} \times 5^{2} +^{6} C_{1} \times^{3} C_{2} \times 5}{6^{4}}$
$\frac{6 (1 + 75 + 15)}{6^{4}} = \frac{91}{216}$

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Four fair dice D1,D2,D3 and D4 each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously. The probability that D4 shows a number appearing on one of D1,D2 and D3, is ?

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