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Mathematics 29 Online
OpenStudy (anonymous):

a certain tennis player makes a successful first serve 73% of the time. assume that each serve is independent of the others. if she serves 6 times, what's the probability she gets a)all 6 serves in? b) exactly 2 serves in? c) at least 4 serves in? d) no more than 4 serves in?

OpenStudy (kropot72):

a) Use Bernouli's Theorem. \[P(6)=\left(\begin{matrix}6 \\6\end{matrix}\right)(0.73)^{6}=(0.73)^{6}\]

OpenStudy (kropot72):

b) Use Bernouli's Theorem again. \[P(2)=\left(\begin{matrix}6 \\ 2\end{matrix}\right)(0.73)^{2}(1-0.73)^{6-2}= \frac{6\times 5}{2}(0.73)^{2}(0.27)^{4}=?\] Can you calculate the answers to a) and b) ?

OpenStudy (anonymous):

yes i can, so to get all 6 serves in, it would be equal to .1513342263

OpenStudy (anonymous):

and to get exactly 2 serves in would be .0012483502

OpenStudy (anonymous):

wait i made a mistake for the 2 serves one, one second

OpenStudy (anonymous):

actually, i'm a bit confused. where did you get the \[(6x5)/2\] from?

OpenStudy (kropot72):

\[\frac{6\times 5}{2}\] comes from calculation of the binomial coefficient \[\left(\begin{matrix}6 \\ 2\end{matrix}\right)=\frac{6!}{2!(6-2)!}=\frac{6!}{2!4!}=\frac{6\times 5}{2}\]

OpenStudy (anonymous):

but \[6!\] is equal to 6 x 5 x 4 x 3 x 2 x 1, how come it's just 6 x 5 there?

OpenStudy (kropot72):

\[\frac{6\times 5\times 4\times 3\times 2\times 1}{2\times 1\times 4\times 3\times 2\times 1}=\frac{6\times 5}{2}\] Notice how the last 4 terms of factorial 6 cancel with the 4 terms of factorial 4.

OpenStudy (anonymous):

ohhh, i see. thank you so much, you helped alot. :)

OpenStudy (kropot72):

Good! Well done. Shall we continue?

OpenStudy (anonymous):

yes, i'm calculating for the exactly 2 serves question now :)

OpenStudy (anonymous):

so for 2 serves, the probability is .0424807263

OpenStudy (anonymous):

so to find the probability that she gets at least 4 serves, would we need to find the probability of her getting 4 serves, 5 serves, and 6 serves? and add them all up

OpenStudy (kropot72):

Good work :). You have the correct answers for a) and b). Now to tackle c). To find the probability of getting at least 4 first serves in we need to find the probabilities of getting exactly 4 serves in, 5 serves in and six serves in and sum the 3 probabilities. We already have P(6), so we need to find P(4) and P(5) \[P(4)=\left(\begin{matrix}6 \\ 4\end{matrix}\right)(0.73)^{4}(0.27)^{2}=\frac{6\times 5}{2}(0.73)^{4}(0.27)^{2}\] What do you get for P(4) ?

OpenStudy (anonymous):

for P(4)= .3105347653 and for P(5)= .3358375981

OpenStudy (anonymous):

and i add the 3 probabilities together and i get .7977065897

OpenStudy (kropot72):

Excellent work! You are really grasping this very well. All your answers are all correct so far. What are your thoughts on d) ?

OpenStudy (anonymous):

for d, i would find the probabilities for P(0) + P(1) + P(2) + P(3) + P(4)

OpenStudy (kropot72):

That's right. You already have P(2) and P(4). Do you need equations for the others or can you try finding them yourself?

OpenStudy (anonymous):

for P(0) though, when i'm doing the factorial part i get \[(6!)/0!(6-0)!\]

OpenStudy (anonymous):

will i get an undefined answer since 0! is 0 making the denominator 0

OpenStudy (kropot72):

Very good question. For the purposes of statistics 0! has been defined as 1.

OpenStudy (anonymous):

ohh i see. thank you SO SO much, i've been struggling with this in class but i finally understand now. thank you! :)

OpenStudy (kropot72):

That's marvellous. I'm so happy to have been able to help :)

OpenStudy (anonymous):

oh, one last thing. is there a shortcut in simplifying the factorial part right away? for example, how you went from \[6!/2(6-2)!\] to (6x5)/2

OpenStudy (anonymous):

because each time i did it, i wrote out all the number's and then cancelled it when i wrote everything out

OpenStudy (kropot72):

Yes there is a shortcut. 6! = 6 * 5 * 4! If you split up 6! as above you can see right away how dividing by 4! gives 6 * 5

OpenStudy (anonymous):

oh yes. i see it now. wow, makes life a bit simpler haha, thank you. also, sorry to be a bother. when i was calculating the probability for getting 0 serves in, the equation i used was (.73)^0 x (.27)^6

OpenStudy (anonymous):

(.73)^0 = 1 but when i entered (.27)^6 in the calculator, i get 3.87420489 E-4

OpenStudy (anonymous):

i don't quite understand what that E-4 is

OpenStudy (kropot72):

Your calculator is set up to give the answer in scientific notation. 3.87420489 E-4 = 3.87420489 * 10^-4 = 0.000387420489

OpenStudy (kropot72):

E stands for exponent. In this case the exponent of 10.

OpenStudy (anonymous):

ohh, got it. i used a graphing calculator when i got that E-4 but when i used a regular scientific calculator, i got the .000387420489. wow, really, thank you so very much for helping me through all these parts for this problem. i really appreciate it :)

OpenStudy (kropot72):

It's good to clear up all these things now :)

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