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.

- What is the probability that the player sinks three baskets in 10 independent throws?
- What is the probability that the player throws 10 times before obtaining the first basket?
- What is the probability that the player throws 10 times before obtaining two baskets?

Answer

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: $\theta=0.35$

$$P(K=k)={{r-1+k}\choose k}\theta^r(1-\theta )^k$$ $$P(K=8)={{2-1+8}\choose 8}(0.35)^2(1-0.35 )^8$$ $$P(K=8)={9\choose 8}(0.35)^2(0.65 )^8$$ $$P(K=8)=9*(0.35)^2(0.65 )^8$$ $$P(k=8)\approx 0.0351$$