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Manjari Bobal
Asked: 2022-11-08T22:37:21+05:30 In:

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.
Which of the following is not true ?
A. `P_(100)gt2//3`
B. `P_(100)lt2//3`
C. `P_(100),P_(101)gt2//3`
D. None of these

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.
Which of the following is not true ?
A. `P_(100)gt2//3`
B. `P_(100)lt2//3`
C. `P_(100),P_(101)gt2//3`
D. None of these
Parabola
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  1. Correct Answer – C
    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)`

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Deepesh Pratap Acharya
Asked: 2022-11-02T01:27:29+05:30 In:

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)` is equal to
A. `(1//2)[P_(n-1)+P_(n-2)]`
B. `(1//2)[2P_(n-1)+P_(n-2)]`
C. `(1//2)[P_(n-1)+2P_(n-2)]`
D. None of these

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)` is equal to
A. `(1//2)[P_(n-1)+P_(n-2)]`
B. `(1//2)[2P_(n-1)+P_(n-2)]`
C. `(1//2)[P_(n-1)+2P_(n-2)]`
D. None of these
Parabola
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  1. Correct Answer – A
    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)`

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Obaid Shenoy
Asked: 2022-10-30T19:50:23+05:30 In:

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
A. `1//2`
B. `2//3`
C. 1
D. None of these

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
A. `1//2`
B. `2//3`
C. 1
D. None of these
Parabola
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  1. Correct Answer – C
    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)`

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