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Question

A coin is so weighted such that the probability of it showing H(head) is 23and that of T(Tail)is 13. When it is tossed. If head appears, then a number from the first 9 naturals is selected at random, otherwise a number from 1, 2, 3, 4, 5 will be selected. Let E be the event of getting an even number, then


A

P(EH)=49

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B

P(ET)=25

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C

P(E)=58135

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D

P(HE)=2029

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Solution

The correct options are
A

P(EH)=49


B

P(ET)=25


C

P(E)=58135


D

P(HE)=2029


Since there are 4 even number among 1 to 9, it follow that
P(EH)=49
Again, since there are 2 even number among 1 to 5, it follow that
P(ET)=25
Therefore (A) and (B) are correct, now
E=(HT)E=(HE)(TE)
This implies
P(E)=P(HE)+P(TE)=P(H)P(EH)+P(T)P(ET)=23×49+13×25=58135
By Baye’s theorem
P(HE)=P(H)P(EH)P(H)P(EH)+P(T)P(ET)=23×4923×49+13×25=2029


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