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and your questions is? .-.
@Luigi0210 @Nnesha @ParthKohli
@ganeshie8 think she wants you .-.
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
binomial distribution
P(X>=2) = 1 - P(X=0) + P(X=1)
do you know how to find P(X=0), the probabiltiy for getting 0 broken cookies ?
no :/ i dont get this at all
thats ok notice that probability for getting a broken cookie = 0.03 therefore probability for NOT getting a broken cookie = 0.97 yes ?
since we want 0 broken cookies : P(X=0) = 0.97^36
so its 0.33??
thats the probability for getting 0 broken cookies but thats not the answer
oh okay
next find the probabilty for getting 1 broken cookie
ok what is next? how did u find .294?
did you get how we arrived at 0.33 for the probability of getting 0 broken cookies ?
yes sir!
good, similarly see if you can find the probability of getting 1 broken cookie
give it a try
I'm not sure.. i tried but i don't know how to get it
@ganeshie8
1 broken cookie means, 35 unbroken cookies, yes ?
yes
so the probability is 0.03*0.97^35 ?
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
okay so that will be 0.2?
try again, im getting 0.37 something
i just want to know how u got .294
yeah same.. i just rounded it
0.4 i meant sorry
so the probability of getting at least 2 broken cookies is \[\large 1-(.97^{36} + 36*0.03*0.97^{35})\]
okay thank you so much!! appreciate it!
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