IIT-JEE 2012 · Paper 2 · Mathematics Q46

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MathematicsSingle correct+3 / -1Question 46 of 60

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 $D_3$ is

\dfrac{91}{216}

\dfrac{108}{216}

\dfrac{125}{216}

\dfrac{127}{216}

Answer and worked solution

Official answer: A

Complement. $D_4$ fails to match when none of $D_1,D_2,D_3$ shows its number: probability $\left(\frac56\right)^3=\frac{125}{216}$ .

Answer. $1-\frac{125}{216}=\frac{91}{216}$ .

(A)

Topics:Classical probability

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IIT-JEE 2012 · Paper 2, Mathematics question 46