Question

a baseball player's batting average is 0.250. what is the probability that he gets exactly one hit in his next four times at bat?

Answer 1

binomial probability for this one
do you know the formula, or do you want to work it out?

Answer 2

your choice, let me know

Answer 3

you work for it..:P

Answer 4

ok without resorting to a formula

Answer 5

sure

Answer 6

what is te variance ?

Answer 7

batting average is \(.25\) meaning she gets a hit \(\frac{1}{4}\) of the time, i.e. with probability \(\frac{1}{4}\) and therefore does not get a hit with probability \(\frac{3}{4}\)

Answer 8

lets compute the probability that in 4 tries she gets (Hit,not hit, not hit, not hit) i.e. gets a hit on the first at bat and doesn't on the next 3
assuming the events are independent, all we have to do is multiply
\[\frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}\times \frac{3}{4}\] better written as
\[\frac{1}{4}\times (\frac{3}{4})^3\]

Answer 9

but that is not the only way for her to get one hit at 4 at bats
could also get
(not hit, Hit, not hit, not hit)
(not hit, not hit, Hit, not hit) or
(not hit, not hit, not hit, Hit)

in other words there are 4 ways to do this, each with the same probability
since these events are disjoint (they are all different) you add up the probabilities, and since they are all the same adding them is the same as multiplying by 4, so you get
\[4\times \frac{1}{4}\times \left(\frac{3}{4}\right)^3\]

Answer 10

we could have gone straight to the formula
\[P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}\] and put \(n=4, k=1, p=0.25\) to get
\[\dbinom{4}{1}.25^1(.75)^3\]

Answer 11

it is good to know the formula for more complex problems, but for this one we could think and arrive at the same answer

Answer 12

guess we lost @CaptainAlice