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Mathematics 15 Online
OpenStudy (anonymous):

HEEEEEEEEEEEEELPPPP! HELP! HELP!!!!! EMERGENCY!!!!!!!!!!!!!!!!! @Thesmartone @hero @kainui @wio

OpenStudy (anonymous):

and your questions is? .-.

OpenStudy (anonymous):

@Luigi0210 @Nnesha @ParthKohli

OpenStudy (anonymous):

@ganeshie8 think she wants you .-.

OpenStudy (anonymous):

The probability of finding a broken cookie in a bag of chocolate chip cookies is P= .03. Find the probability of getting at least 2 broken cookies in a bag containing 36 cookies. A) .91 B) .294 C) .33 D) .06 @ganeshie8

ganeshie8 (ganeshie8):

binomial distribution

ganeshie8 (ganeshie8):

P(X>=2) = 1 - P(X=0) + P(X=1)

ganeshie8 (ganeshie8):

do you know how to find P(X=0), the probabiltiy for getting 0 broken cookies ?

OpenStudy (anonymous):

no :/ i dont get this at all

ganeshie8 (ganeshie8):

thats ok notice that probability for getting a broken cookie = 0.03 therefore probability for NOT getting a broken cookie = 0.97 yes ?

ganeshie8 (ganeshie8):

since we want 0 broken cookies : P(X=0) = 0.97^36

OpenStudy (anonymous):

so its 0.33??

ganeshie8 (ganeshie8):

thats the probability for getting 0 broken cookies but thats not the answer

OpenStudy (anonymous):

oh okay

ganeshie8 (ganeshie8):

next find the probabilty for getting 1 broken cookie

OpenStudy (anonymous):

ok what is next? how did u find .294?

ganeshie8 (ganeshie8):

did you get how we arrived at 0.33 for the probability of getting 0 broken cookies ?

OpenStudy (anonymous):

yes sir!

ganeshie8 (ganeshie8):

good, similarly see if you can find the probability of getting 1 broken cookie

ganeshie8 (ganeshie8):

give it a try

OpenStudy (anonymous):

I'm not sure.. i tried but i don't know how to get it

OpenStudy (anonymous):

@ganeshie8

ganeshie8 (ganeshie8):

1 broken cookie means, 35 unbroken cookies, yes ?

OpenStudy (anonymous):

yes

ganeshie8 (ganeshie8):

so the probability is 0.03*0.97^35 ?

ganeshie8 (ganeshie8):

but wait, that broken cookie could be either 1st or 2nd or 3rd or... 36th there are 36 ways it can happen so the probability for getting 1 broken cookie is actually 36* 0.03*0.97^35

OpenStudy (anonymous):

okay so that will be 0.2?

ganeshie8 (ganeshie8):

try again, im getting 0.37 something

OpenStudy (anonymous):

i just want to know how u got .294

OpenStudy (anonymous):

yeah same.. i just rounded it

OpenStudy (anonymous):

0.4 i meant sorry

ganeshie8 (ganeshie8):

so the probability of getting at least 2 broken cookies is \[\large 1-(.97^{36} + 36*0.03*0.97^{35})\]

OpenStudy (anonymous):

okay thank you so much!! appreciate it!

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