A six faced die is a biased one. It is thrice more likely to show an odd number than to show an even number. It is thrown twice. The probability that the sum of the numbers in the two throws is even is
Step-by-step Solution:
Let the probability of showing an even number be \( P \). According to the question, the probability of showing an odd number is \( 3P \). From the total probability: \[ 3P + P = 1 \] Solving: \[ P = \frac{1}{4}, \quad 3P = \frac{3}{4} \] Now, the sum of the numbers in two throws is even when either both numbers are odd or both are even. The required probability: \[ = \frac{3}{4} \times \frac{3}{4} + \frac{1}{4} \times \frac{1}{4} \] \[ = \frac{9}{16} + \frac{1}{16} = \frac{10}{16} = \frac{5}{8} \]