JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Probability

When a certain biased die is rolled, a particular face occurs with probability

\frac{1}{6} - x

and its opposite face occurs with probability

\frac{1}{6} + x

. All other faces occur with probability

\frac{1}{6}

. Note that opposite faces sum to 7 in any die. If

0 < x < \frac{1}{6}

, and the probability of obtaining total sum

= 7

, when such a die is rolled twice, is

\frac{13}{96}

, then the value of x is:

Answer: (C)

Solution

Let the biased opposite pair carry probabilities

\frac{1}{6} - x

and

\frac{1}{6} + x

, the other two opposite pairs each

\frac{1}{6}, \frac{1}{6}

. A total of

7

requires opposite faces on the two throws, so summing over the three opposite pairs and both orders,

P = 2 (\frac{1}{6} - x) (\frac{1}{6} + x) + 2 \cdot \frac{1}{6} \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{6} - 2 x^{2} .

Setting

\frac{1}{6} - 2 x^{2} = \frac{13}{96}

gives

2 x^{2} = \frac{16 - 13}{96} = \frac{1}{32}

, so

x^{2} = \frac{1}{64}

. With

0 < x < \frac{1}{6}

x = \frac{1}{8}

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Single correct option