A player tosses a coin and score one point for every head and two points for every tail that turns up. He plays on until his score reaches or passes n. `P_(n)` denotes the probability of getting a score of exactly n.
The value of `P_(n)+(1//2)P_(n-1)` is equal to

Options

A. (a) `1//2`
B. (b) `2//3`
C. (c) 1
D. (d) None of these

Answer

C

Explanation

The scores of n can be reached in the following two mutually exclusive events:
(i) by throwing a head when the score is `(n-1),`
(ii) by throwing a tail when the score is `(n-2)`
Hence `P_(n)=P_(n-1)xx1/2+P_(n-2)+1/2" "[becauseP("head")="(tail")=1//2"]"`
`=1/2[P_(n-1)+P_(n-2)]" "(1)`
` impliesP_(0)+1/2P_(n-1)=P_(n-1)+1/2P_(n-2)`
`" "("adding"(1//2)P_(n-1)"on both sides")`
`=P_(n-2)+1/2P_(n-3)`
`=P_(2)+1/2P_(1)" "(2)`
Now, a score of 1 can be obtained by throwing a head at a single toss. Therefore, `P_(1)=1/2`
And a score of 2 can be obtained by throwing either a tail at a single toss or a head at the first toss as well as second toss. Therefore,
`P_(2)=1/2+((1)/(2)xx(1)/(2))=3/4`
From Eq. (2), we have
`P_(n)+1/2P_(n-1)=3/4+1/2((1)/(2))=1`
`or P_(n)=1-1/2P_(n-1)`
`or P_(n)-2/3=1-1/2P_(n-1)-2/3`
`or P_(n)-2/3=-1/2(P_(n-1)(2)/(3))`
`=(-(1)/(2))^(2)(P_(n-1)-(2)/(3))`
`=(-(1)/(2))^(3)(P_(n-3)-(2)/(3))`
`=(-(1)/(2))^(n-1)(P_(1)-(2)/(3))`
`=(-(1)/(2))^(n-1)((1)/(2)-(2)/(3))`
`=(-(1)/(2))^(n-1)(-(1)/(6))`
`(-(1)/(2))^(n)1/3`
`or P_(n)=2/3+((-1)^(n))/(2^(n))1/3=1/3{2+((-1)^(n))/(2^(n))}`
Now, `P_(100)=2/3+(1)/(3xx2^(101))gt2/3`
and `P_(101)=2/3-(1)/(3xx2^(101))lt2/3`
`P_(101)lt2/3ltP_(100)`