Question

A six-faced die is a biased one. It is three times more likely to show an odd number than an even number. It is thrown twice. The probability that the sum of the numbers in two throws is even is:

• A. \( \frac{5}{9} \)
• B. \( \frac{5}{8} \)
• C. \( \frac{1}{2} \)
• D. None of these

Hint:

When dealing with biased dice, carefully use the given odds to set up probabilities and then calculate the likelihood of the event (like even sums) by considering all possible outcomes.

Answer 1

The Correct Option is B

Let the probability of getting an odd number be \( p \) and the probability of getting an even number be \( q \). We are told that it is three times more likely to show an odd number than an even number, so: \[ p = 3q \] Since the die has only odd and even numbers, the sum of the probabilities must equal 1: \[ p + q = 1 \] Substitute \( p = 3q \) into the equation: \[ 3q + q = 1 \quad \Rightarrow \quad 4q = 1 \quad \Rightarrow \quad q = \frac{1}{4} \] Now, substitute \( q = \frac{1}{4} \) into \( p = 3q \): \[ p = 3 \times \frac{1}{4} = \frac{3}{4} \] Thus, the probability of getting an odd number is \( \frac{3}{4} \), and the probability of getting an even number is \( \frac{1}{4} \). Step 1: Probability of getting an even sum The sum of the numbers will be even if both numbers are either odd or even. Therefore, we calculate the probabilities of these two events: 1. The probability of getting an odd number on both throws is: \[ {Probability of odd and odd} = p \times p = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16} \] 2. The probability of getting an even number on both throws is: \[ {Probability of even and even} = q \times q = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \] Thus, the total probability of getting an even sum is: \[ {Total probability} = \frac{9}{16} + \frac{1}{16} = \frac{10}{16} = \frac{5}{8} \] Therefore, the probability that the sum of the numbers in two throws is even is \( \frac{5}{8} \), making the correct answer Option B.