Binomial Probability Problem

• May 9th 2011, 07:59 AM
math321
Binomial Probability Problem
hey guys i got a final exam on wednesday and i am working on some practice questions
jus need some help in doing this 1 dont really care about the answer
i jus want the steps in solving this

every now and again even a good diamond cutter has a problem and the diamond breaks for one particular cutter the rate of breakage is 0.1
i. give reasons why the binomial probability model is suited for this problem
if the cutter works on 75 stones
a. what is the probability that he breaks exactly 3 stones
b. what is the probability that he breaks 2 or more stones
c. find the mean of this distribution
d. find the stand dev of this distribution
• May 9th 2011, 08:19 AM
TheEmptySet
Quote:

Originally Posted by math321
hey guys i got a final exam on wednesday and i am working on some practice questions
jus need some help in doing this 1 dont really care about the answer
i jus want the steps in solving this

every now and again even a good diamond cutter has a problem and the diamond breaks for one particular cutter the rate of breakage is 0.1
i. give reasons why the binomial probability model is suited for this problem
if the cutter works on 75 stones
a. what is the probability that he breaks exactly 3 stones
b. what is the probability that he breaks 2 or more stones
c. find the mean of this distribution
d. find the stand dev of this distribution

The answers to c and d can be found here
Binomial distribution - Wikipedia, the free encyclopedia

The reason is that these are Bernoulli trials there are exactly two possible outcomes the diamond breaks or it doesn't.

$P(X=x)=\binom{75}{x}(0.1)^x(0.9)^{75-x}$

Hint for b use the compliment of the event.
• May 9th 2011, 08:24 AM
Sambit
i. Binomial probability model is applicable whenever the outcome of a random experiment (here cutting of a diamond) can be classified into 2 categories: success (here the breaking of diamond) and failure (not breaking). So it is applicable here.

P.S.: Don't get confused with the practical meaning of "success" and "failure". You are to find out probability of breaking, so we take "probability of breaking" as success.

a. If $X$ is the number of diamonds that break amongst the 75 diamonds, then $X\sim Bin(n=75,p=0.1)$.

Then $P(X=x) = {{n}\choose{x}}p^x(1-p)^{n-x}$

So the probability that he breaks exactly 3 stones = $P(X=3)$

b. the probability that he breaks 2 or more stones = $P(X\geq 2) = 1-P(X<2) = P(X=0) + P(X=1)$

c. mean = $np$

d. standard deviation = $\sqrt{np(1-p)}$
• May 9th 2011, 10:51 AM
math321
so when is not applicable
• May 9th 2011, 12:30 PM
Plato
Quote:

Originally Posted by math321
so when is not applicable

Quote:

Originally Posted by math321
so when is not applicable

Essentially, the trials must be independent.
If the outcomes are not independent it is not applicable.
For example, if we deal five cards the outcomes are not independent.
On the other hand, if we draw one card at a time returning the card to the deck each time then the outcomes are independent.
• May 9th 2011, 12:47 PM
math321
cud u check to see if p=.001 or .1
• May 9th 2011, 01:05 PM
mr fantastic
Quote:

Originally Posted by math321
cud u check to see if p=.001 or .1

Ask whoever gave you these practice questions for the solutions or, at least, the answers. (It is not our job to do the work of your teacher).
• May 9th 2011, 01:11 PM
Plato
Quote:

Originally Posted by math321
cud u check to see if p=.001 or .1

The way you have written the question it says 0.1.
However, that is 10%. That is high failure rate.
Perhaps there is something about the wording that implies 0.001.
That is 0.1% which is more like it.
BUT as written it is 0.1.
• May 9th 2011, 01:13 PM
math321
yep the question says .1% so i made a typo
working on it right now thanks though
• May 9th 2011, 01:26 PM
math321
got this when i solved
P (X = 3) = \binom{75}{3} . 0.001^3 ((1-0.001)^(75-3))
• May 9th 2011, 01:50 PM
math321
for my final answer i got .000062832
cud u verify that for me
• May 9th 2011, 08:47 PM
Sambit
Quote:

Originally Posted by math321
for my final answer i got .000062832
cud u verify that for me

Yes $P(X=3)=.000062832$ if $p=.001$