Question:
A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find P(G), where G is the event that a number greater than 3 occurs on a single roll of the die.
Solution:
Given that probability of odd numbers
= 2 × (Probability of even number)
⇒ P (Odd) = 2 × P (Even)
Now, P (Odd) + P (Even) = 1
⇒ 2 P (Even) + P (Even) = 1
⇒ 3 P (Even) = 1
P (Even) = 1/3
So,
$\mathrm{P}($ Odd $)=1-\frac{1}{3}=\frac{3-1}{3}=\frac{2}{3}$
Now, Total number occurs on a single roll of die = 6
And the number greater than 3 = 4, 5 or 6
So, P (G) = P (number greater than 3)
= P (number is 4, 5 or 6)
Here, 4 and 6 are even numbers and 5 is odd
∴ P (G) = 2 × P (Even) × P (Odd)
= 2 × 1/3 × 2/3
= 4/9
Hence, the required probability is 4/9