Chapter 2 Section 3 Question 15
Question
Suppose that a basketball player sinks a basket from a certain position on the court with probability 0.35.
1. What is the probability that the player sinks three baskets in 10 independent throws?
2. What is the probability that the player throws 10 times before obtaining the first basket?
3. What is the probability that the player throws 10 times before obtaining two baskets?
Author: Mohammad-Ali Bandzar| Date:Oct 16 2020
(a) What is the probability that the player sinks three baskets in 10 independent throws?
This question is asking for the binomial probability as all our trials are independent.
From Example 2.3.3 in the book we know: $$p\scriptsize{Y}\normalsize{(y)=P(Y=y)={n\choose y}\theta ^y(1-\theta)^{n-y}}$$ n represents the number of trials
$\theta$ represents the probability of a success
y represents the desired number of successful outcomes

In this question the probability of success is given to us, therefore: $\theta=0.35$
We have 10 independent trials. Therefore $n=10$
$$Y ~ Binomial(10, 0.35)$$ $$P(Y=y)={n\choose y}\theta ^y(1-\theta)^{n-y}$$ $$P(Y=3)={10\choose 3}(0.35) ^3(1-0.35)^{10-3}$$ $$P(Y=3)=120(0.35) ^3(0.65)^{7}$$ $$P(Y=3)\approx 0.2522$$
(b) What is the probability that the player throws 10 times before obtaining the first basket?
Since we are trying to calculate number of failures before the first success we have a geometric distribution
From Example 2.3.4 in the book we know: $$p\scriptsize{X}\normalsize{(x)=P(X=x)=(1-\theta )^x\theta}$$ $\theta$ is the probability of success per event
x represents the number of failed events before a success.

in this case we want 9 failures before a success so $x=9$
In this question the probability of success is given to us, therefore: $\theta=0.35$
$$P(X=9)=(1-0.35 )^9(0.35)$$ $$P(X=9)=(0.65 )^9(0.35)$$ $$P(X=9)\approx 0.072$$
(c) What is the probability that the player throws 10 times before obtaining two baskets?
This is negative binomial distribution as our trials stop after a certain number of successful events. Since the desired number of successful events is greater than one this is negative-Binomial distribution instead of geometric.
From Example 2.3.5 in the book we know: $$p\scriptsize{K}\normalsize{(k)=P(K=k)={{r-1+k}\choose k}\theta^r(1-\theta )^k}$$ r represents the desired number of successes
k represents the number of failures
$\theta$ represents the probability of a success.

since we want two successes $r=2$
since we want 10 total trials, trials-successes=failures: $k=10-2=8$ In this question the probability of success is given to us, therefore: $\theta=0.35$
$$P(K=k)={{r-1+k}\choose k}\theta^r(1-\theta )^k$$ $$P(K=8)={{2-1+8}\choose 8}(0.35)^2(1-0.35 )^8$$ $$P(K=8)={9\choose 8}(0.35)^2(0.65 )^8$$ $$P(K=8)=9*(0.35)^2(0.65 )^8$$ $$P(k=8)\approx 0.0351$$