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Question

A player tosses a coin and scores one point for every head and two point for every tail that truns up. He plays on until his scores reaches or psses n. Pn denotes the probability of getting a scores of exactly n

List IList II(a) the value of Pn is (p) 1(b) the value of Pn+12Pnâˆ’1(q) 54(c) 2P101+P100(r) 2(d) P1+P2(s) 12[Pnâˆ’1+Pnâˆ’2]

Which of the following is the onlycorrect option?

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Solution

The correct option is **A** (a)→(s)

The scores of n can be reached in the following two mutually exlcusive events

a:by throwing a head when the score is (n−1)

b:by throwing a tail when the score is (n−2)

Pn=12[Pn−1+Pn−2]

Pn+12Pn−1=Pn−1+12Pn−2⇒Pn+12Pn−1=P2+12P1

Now P1=12

For P2=12+12×12=34→Pn+12Pn−1=34+12×12=1

Pn−23=1−12Pn−1−23=−12(Pn−1−23)→Pn−1−23=−12(Pn−2−23)∴Pn−23=1(−2)n−1(P1−23)Pn=23+(−1)n2n×3P100=23+13×2100P101=23−13×2101P100+2P100=2

The scores of n can be reached in the following two mutually exlcusive events

a:by throwing a head when the score is (n−1)

b:by throwing a tail when the score is (n−2)

Pn=12[Pn−1+Pn−2]

Pn+12Pn−1=Pn−1+12Pn−2⇒Pn+12Pn−1=P2+12P1

Now P1=12

For P2=12+12×12=34→Pn+12Pn−1=34+12×12=1

Pn−23=1−12Pn−1−23=−12(Pn−1−23)→Pn−1−23=−12(Pn−2−23)∴Pn−23=1(−2)n−1(P1−23)Pn=23+(−1)n2n×3P100=23+13×2100P101=23−13×2101P100+2P100=2

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