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A die is loaded in such a way that a way...

A die is loaded in such a way that 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.

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It is given that, `2xx` Probability of even number = Probability of odd number
`rArr" "P(O)=2P(E)`
`rArr" "P(O):P(E)=2:1`
`"Probability of occurring odd number, " P(O)=(2)/(2+1)=(2)/(3)`
`"and probability of occurring 5each number, " P(E)=(1)/(2+1)=(1)/(3)`
Now, G be the even that a number greater than 3 occur in a single roll of die.
So, the possible outcomes are 4, 5 and 6 out of which two are even and one odd.
`therefore" " "Required probability"=P(G)=2xxP(E)xxP(O)`
`=2xx(1)/(3)xx(2)/(3)=(4)/(9)`
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