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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 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`.
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