When a certain biased die is rolled, a particular face occurs with probability 1/6 - x and its opposite face occurs with probability 1/6 + x. All other faces occur with probability 1/6. Note that opposite faces sum to 7 in any die. If 0 < x < 1/6, and the probability of obtaining total sum = 7, when such a die is rolled twice, is 13/96, then the value of x is :

Solution

Maths sir · 4 steps

Only the two outcomes that use the biased faces change; the x terms cancel, leaving just -2x².

1. List the pairs. A total of 7 comes from six ordered outcomes: (1,6), (6,1), (2,5), (5,2), (3,4), (4,3). Each pair is one face and its opposite face. Call the biased faces a and b, with P(a) = 1/6 - x and P(b) = 1/6 + x.
2. Biased pair. The outcomes (a,b) and (b,a) each have probability (1/6 - x)(1/6 + x) = 1/36 - x². The other four outcomes use fair faces, so each has probability 1/36.

   P(sum = 7) = 2(1/36 - x²) + 4/36 = 1/6 - 2x²

3. Solve for x. Set 1/6 - 2x² = 13/96. Since 1/6 = 16/96, this gives 2x² = 3/96 = 1/32, so x² = 1/64.
4. So. x = 1/8. It lies between 0 and 1/6, so it is allowed. The answer is (b).

The idea to keep. For a sum on two dice, group the outcomes into pairs. When the biases are +x and -x on opposite faces, the product is 1/36 - x², so only x² survives.

Answer from the NTA final answer key. Solution by PrepWiser.