Question of the Week- 20 (A biased dice is such that any odd number..)

Solution

• In a biased dice, any odd number appears thrice as frequently as any even number

• The probability that the sum of the numbers obtained upon rolling the dice twice is 10

• Let us assume that the frequency of getting any even number, {2, 4, 6} is x

o Then, the frequency of getting any odd number, {1, 3, 5} will be 3x

• Implies,

o The probability of getting any even number = \(\frac{x}{(x + 3x + x + 3x + x + 3x)} = \frac{1}{12}\)
o And, the probability of getting any odd number = \(\frac{3x}{(x + 3x + x + 3x + x + 3x)} = \frac{3}{12} = \frac{1}{4}\)

• For the sum of two outcomes to be 10, the outcomes must be either (4, 6) or (5, 5) or (6, 4)
• Therefore, P(sum = 10) = P(4, 6) + P(5, 5) + P(6, 4) = \((\frac{1}{12} * \frac{1}{12}) + (\frac{1}{4} * \frac{1}{4}) + (\frac{1}{12} * \frac{1}{12}) = \frac{11}{144}\)