SOLUTION: in a box of 100 chocolate bars, there are 20 Kit Kat, 30 smarties, 10 coffees crisps and 40 aeros. You draw a chocolate bars and replace it, what is the probability of drawing atle

Algebra ->  Probability-and-statistics -> SOLUTION: in a box of 100 chocolate bars, there are 20 Kit Kat, 30 smarties, 10 coffees crisps and 40 aeros. You draw a chocolate bars and replace it, what is the probability of drawing atle      Log On


   



Question 1179724: in a box of 100 chocolate bars, there are 20 Kit Kat, 30 smarties, 10 coffees crisps and 40 aeros. You draw a chocolate bars and replace it, what is the probability of drawing atleast 3 smarties
Found 2 solutions by ikleyn, Theo:
Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
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in a box of 100 chocolate bars, there are 20 Kit Kat, 30 smarties, 10 coffees crisps and 40 aeros.
You draw a chocolate bars and replace it, what is the probability of drawing atleast 3 smarties
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As it is worded,  printed,  posted and presented,  the problem is posed  INCORRECTLY.


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@Theo incorrectly interprets and solves the problem.


In order for the problem be posed correctly, the number of trials should be given in the condition -

                BUT IT IS NOT GIVEN IN THE PROBLEM.

    Therefore, the "solution" by @Theo is a FICTION, is INCORRECT, is FALSE and EMPTY.



Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
the probability of getting at lest 3 is equal to 1 minus the probability of getting less than 3.

the probability of getting less than 3 is equal to the probability of getting 0 or 1 or 2 only.

this is a binomial problem where p(x) = c(n,x) * p^x * q^(n-x).
n = 100
x = 0 or 1 or 2
c(n,x) = n! / (x! * (n-x)!
specifically, ...

p(one smartie) = 30/100 = 3/10 = .3
p(anything but one smartie) = 1 - .3 = .7

since you are replacing the chocolate bar each time, the probability remains the same for each draw.

p(0) = .3^0 * .7^(100 - 0) * c(100,0) = 3.23447651 * 10^-16
p(1) = .3^1 * .7^(100 - 1) * c(100,1) = 1.38620422 * 10^-14
p(2) = .3^2 * .7^(100 - 2) * c(100,2) = 2.94073323 * 10^-13

the total probability of getting less than 3 is equal to 3.08258813 * 10^-13.

the total probability of getting at least 3 is 1 minus that.

my calculator couldn't handle it, so i went to excel.

this is what i got.

p(less than 3) = 3.08258813312174 * 10^-13

p(3 or greater) = 1 minus p(less than 3) = 9.99999999999686 * 10^-1

i went a little further.

the best excel could give me is:

p(3 or greater) = 0.999999999999692

i resorted to calculating it manually and the best that i could do is:

p(3 or greater) = 0.999999999999601741186687826

it's kind of like a moot point.

if you round your answer to anything less than 12 digits, the answer is p(3 or greater) = 1.

sticking to the answer that the calculator gave me, i would go with p(3 or greater) = 1 or p(3 or greater) = 1 minus p (less than 3) = 1 minus 3.08258813 * 10^-13.