JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Probability XII

Four fair dice

D_{1}, D_{2}, D_{3}

and

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}

and

D_{3}

Answer: (A)

Solution

The complementary event is that

D_{4}

shows a number different from all of

D_{1}, D_{2}, D_{3}

.

For any outcome of

D_{1}, D_{2}, D_{3}

, there are at most 3 distinct values shown. So

D_{4}

has at least 3 values it can show that differ from all three dice.

Actually, the probability that

D_{4}

differs from all three is: if

D_{1}, D_{2}, D_{3}

show values (possibly repeated),

D_{4}

must avoid those values. In the worst case (all three show different values),

D_{4}

has 3 choices out of 6.

The probability that

D_{4}

shows a value different from all of

D_{1}, D_{2}, D_{3}

\frac{6 \cdot 5 ^{3}}{6 ^{4}} = \frac{125}{216}

.

Therefore, the required probability is

1 - \frac{125}{216} = \frac{91}{216}

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Single correct option