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Mathematics 19 Online
OpenStudy (anonymous):

Assuming that Caleb makes 70% of his free throws and shoots 10 free throws in a game, what is the probability that he will make? a) All 10 of his free throws in a game? b) Make seven or more of his free throws in a game? c) Find the mean number of free throws Caleb would make in the game.

OpenStudy (anonymous):

I think i have A, i got 0.0282475249

OpenStudy (anonymous):

@jim_thompson5910 could you help please?

pooja195 (pooja195):

You can solve this using a binomial distribution

pooja195 (pooja195):

do you know how to do that?

OpenStudy (anonymous):

not really I'm confused on that :/

pooja195 (pooja195):

we know the probability of making any given free throw is .70

OpenStudy (anonymous):

yes

pooja195 (pooja195):

S = makes a free throw. F = fails to make a free throw

OpenStudy (anonymous):

okay

pooja195 (pooja195):

a) we want the probability SSS SSS SSS SS , ten S's in a row each S has probability of 0.7 so you get 0.7 * 0.7 * 0.7 * ... *0.7 = 0.7^10

OpenStudy (anonymous):

= .0282475249

pooja195 (pooja195):

yep

OpenStudy (anonymous):

okay so what about b?

pooja195 (pooja195):

b) you want seven successful free throws for example: SSSSSSSFFFF

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

so do we do .7^7 + .7^3?

pooja195 (pooja195):

so we also need eight S's, nine S's , and ten S's

OpenStudy (anonymous):

0.6496107184?

pooja195 (pooja195):

for the case of exactly seven free throws SSS SSS S FFF .7^7 * .3^3 and you're not done, there are also more cases the order counts, you can have FFF SSS F SSSS

pooja195 (pooja195):

b) is asking P( X >= 7 ) = P( X = 7 ) + P( X= 8) + P( X = 9) + P( X = 10)

pooja195 (pooja195):

yep

OpenStudy (anonymous):

okay, c?

pooja195 (pooja195):

im not sure about part C @paki

OpenStudy (anonymous):

@paki can you help?

OpenStudy (paki):

@Nnesha help...

OpenStudy (anonymous):

can you help? @Luigi0210

OpenStudy (anonymous):

@TheSmartOne

pooja195 (pooja195):

@KingGeorge

OpenStudy (anonymous):

thank you @pooja195

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

part c) mean = n*p n = number of trials p = probability of success

OpenStudy (anonymous):

7?

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

yep n*p = 10*0.7 = 7

OpenStudy (anonymous):

okay so can you double check please a) .0282475249 b) 0.6496107184 c) 7

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