A player tosses an unbiased coin, and as a result, he can score 2 points for every head which turns up, and one point for every tail which turns up. The probability of his scoring exactly 15 points =
The correct option is A 23−13(12)15
Let pn denote the probability of scoring exactly n points in an infinite sequence of tosses. Then pn is the answer to our problem. We will set up a recurrence relation for pn using pn−1.
In an infinite sequence of tosses, note that there are only two possibilities:
Case 1:
We scored exactly n−1 points at some point in time. This happens with probability pn−1, by definition. Then if we consider the toss after that, there is a 12 probability we score n points and a 12 probability we score n+1 points. In this case there we land on n with probabilitiy 12 .
Case 2:
We never score exactly n−1 points in any point in time. This happens with probability 1−pn−1. This case implies that the score jumped from n−2 to n, meaning we must have reached n points at a certain point (i.e. probability 1)
Using the law of total probability, our recurrence is then:
pn=(pn−1)⋅12+(1−pn−1)⋅1=−12pn−1+1
This is a linear nonhomogeneous recurrence relation with constant coefficients, which we can use many standard techniques to solve. One method is to transform our recurrence to a homogenous one by subtracting two equations.
If we substitute n→n−1 in our recurrence, we get
pn−1=−12pn−2+1
Subtracting this equation from our original recurrence yields
pn−pn−1=−12pn−1+12pn−2
2pn−pn−1−pn−2=0
Now that our recurrence is homogenous, we can get the corresponding characteristic polynomial by substituting pi with λi, as follows:
2λn−λn−1−λn−2=0
(dividing by λn−2)
2λ2−λ−1=0
(2λ+1)(λ−1)=0
The roots of our characteristic polynomial are λ1=−12 and λ2=1, so the solution to our recurrence is of the form
pn=Aλn1+Bλn2=A(−12)n+B
The last thing to do is to solve for our constants using the initial conditions p0=1 and p1=12.
p0=A+B=1
p1=−\dfrac{1}{2}A+B=12
A=13
B=23
Therefore the probability of scoring exactly n points within a sequence of n tosses is
pn=13(−12)n+23=23+(−12)n13
now put n=15 we will get our desired result,ie option A
One interesting thing to note is that as n→∞, pn→23
Let pn denote the probability of scoring exactly n points in an infinite sequence of tosses. Then pn is the answer to our problem. We will set up a recurrence relation for pn using pn−1.
In an infinite sequence of tosses, note that there are only two possibilities:
Case 1:
We scored exactly n−1 points at some point in time. This happens with probability pn−1, by definition. Then if we consider the toss after that, there is a 12 probability we score n points and a 12 probability we score n+1 points. In this case there we land on n with probabilitiy 12 .
Case 2:
We never score exactly n−1 points in any point in time. This happens with probability 1−pn−1. This case implies that the score jumped from n−2 to n, meaning we must have reached n points at a certain point (i.e. probability 1)
Using the law of total probability, our recurrence is then:
pn=(pn−1)⋅12+(1−pn−1)⋅1=−12pn−1+1
This is a linear nonhomogeneous recurrence relation with constant coefficients, which we can use many standard techniques to solve. One method is to transform our recurrence to a homogenous one by subtracting two equations.
If we substitute n→n−1 in our recurrence, we get
pn−1=−12pn−2+1
Subtracting this equation from our original recurrence yields
pn−pn−1=−12pn−1+12pn−2
2pn−pn−1−pn−2=0
Now that our recurrence is homogenous, we can get the corresponding characteristic polynomial by substituting pi with λi, as follows:
2λn−λn−1−λn−2=0
(dividing by λn−2)
2λ2−λ−1=0
(2λ+1)(λ−1)=0
The roots of our characteristic polynomial are λ1=−12 and λ2=1, so the solution to our recurrence is of the form
pn=Aλn1+Bλn2=A(−12)n+B
The last thing to do is to solve for our constants using the initial conditions p0=1 and p1=12.
p0=A+B=1
p1=−\dfrac{1}{2}A+B=12
A=13
B=23
Therefore the probability of scoring exactly n points within a sequence of n tosses is
pn=13(−12)n+23=23+(−12)n13
now put n=15 we will get our desired result,ie option A
One interesting thing to note is that as n→∞, pn→23
