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.

Answer 1

Given that probability of even numbers

= `1/2` × probability of odd numbers

⇒ P(Odd): P(Even) = 2:1

∴ P(odd number) = `2/(2 + 1) = 2/3`

And P(even number) = `1/(2 + 1) = 1/3`

Also given that, G the event that a number greater than 3 occurs in a single throw of die.

∴ The possible outcome are 4, 5 and 6 out of which two are even and one is odd.

∴ Required probability = P(G)

= 2 × P(even) × P(odd)

= `2 xx 1/3 xx 2/3 = 4/9`

Hence, the required probability is `4/9`.