# Thread: probability: free throw shooter

1. ## probability: free throw shooter

An excellent free throw shooter attempts several free throws until she misses.
a) If p=0.9 is her probability of making a free throw, what is the probability of having the first miss on the 13th attempt or later?
b) If she continues shooting until she misses three, what is the probability that the third miss occurs on the 30th attempt?

Thank you so much in advance for your help... I thought calculus based statistics would make more sense to me.

2. Originally Posted by calabrone
An excellent free throw shooter attempts several free throws until she misses.
a) If p=0.9 is her probability of making a free throw, what is the probability of having the first miss on the 13th attempt or later?
b) If she continues shooting until she misses three, what is the probability that the third miss occurs on the 30th attempt?

Thank you so much in advance for your help... I thought calculus based statistics would make more sense to me.
Are you aware of Geometric and Negative Binomial r.v's? Look up their pdf's and use them as follows.

a) Is a Geometric r.v. with p = 0.1, because we are interested in a miss. Let x be the number of shots attempted. So we want $\displaystyle P (x \geq 13)$.

b) Is a Negative Binomial r.v. with r = 3, and p = 0.1. Again we're interested in the miss. Let x be the number of shots attempted, we want $\displaystyle P ( x = 30).$

3. Is the first one to be done with the following equation?

1-[P(x=12)+P(x=11)+....+P(x=0)]?

I am having such a problem doing this type of equation correctly.

For part b, I'm not sure how to begin. Do you multiply the probability of a miss by 30?

4. Originally Posted by calabrone
Is the first one to be done with the following equation?

1-[P(x=12)+P(x=11)+....+P(x=0)]?

I am having such a problem doing this type of equation correctly.

For part b, I'm not sure how to begin. Do you multiply the probability of a miss by 30?
For a) that is certainly one way of finding the probability. Not sure what you mean by doing the equation correctly. You just plug in the numbers into the formula for a geometric distribution: $\displaystyle (1 - p)^{x - 1}p$. And Do $\displaystyle P(x \geq 13) = 1 - \sum_{x = 1}^{12}(1 - p)^{x - 1}p$ Here p = 0.1 as I said before, since we are interested in the miss. Note that the minimum value for x is not 0, it is 1, because we must shoot at least once to obtain a miss.

Negative Binomial Distr. Formula:$\displaystyle p(x) =$ $\displaystyle x - 1 \choose r - 1$$\displaystyle p^r(1 - p)^{ x -r}\ where\ x \in \{r, r + 1, \ldots\}$
b) I've given you the formula for a negative binomial distribution this time(I asked you to look it up last time). As I said before then using that formula you have r = 3, and p = 0.1. And we want $\displaystyle P(X = 30).$

5. Originally Posted by calabrone
An excellent free throw shooter attempts several free throws until she misses.
a) If p=0.9 is her probability of making a free throw, what is the probability of having the first miss on the 13th attempt or later?
b) If she continues shooting until she misses three, what is the probability that the third miss occurs on the 30th attempt?

Thank you so much in advance for your help... I thought calculus based statistics would make more sense to me.