# Thread: Calculating probability when sampling without replacement.

1. ## Calculating probability when sampling without replacement.

I have two questions which i need some help and tell me what is wrong with my attempted answers:
A box contains 100 items, 2 of which are faulty. If 2 are selected at random without replacement calculate the probabilities that 0, 1 or 2 of the selected items are faulty.

This my attempt for the first probaility:
Pr(0 Faulty)= (99/100) * (98/100)

An electronic components manufacturer submits tenders to three separate companies for a work over the next month. From past experience the contracts manager estimates that the probability that each tender will be accepted as in the table below.
Company Probability
A 0.8
B 0.6
C 0.9

Calculate the probability that at least one tender is accepted:

This is my attempt:
I have assumed this is independent.
Pr(at least one tender) = Pr(A) + Pr(B) + Pr(C) + Pr(A U B) + Pr(A U C) + Pr(B U C) + Pr(A U B U C)

P.S

2. ## Re: Probability questions

Originally Posted by Paymemoney
A box contains 100 items, 2 of which are faulty. If 2 are selected at random without replacement calculate the probabilities that 0, 1 or 2 of the selected items are faulty.
Let $\displaystyle F$ be the number of faulty items in the sample.
Then .

To answer this question sum k=0,1,2.

Did you get 1 as the answer?

3. ## Re: Calculating probability when sampling without replacement.

is that sum P(F = k) calculated as matrices?

because i do not get 1 as the answer

4. ## Re: Calculating probability when sampling without replacement.

Originally Posted by Paymemoney
I have two questions which i need some help and tell me what is wrong with my attempted answers:
A box contains 100 items, 2 of which are faulty. If 2 are selected at random without replacement calculate the probabilities that 0, 1 or 2 of the selected items are faulty.

This my attempt for the first probaility:
Pr(0 Faulty)= (99/100) * (98/100)
This can't be right. There are 100 items but 2 are faulty so there are 98, out of 100, that are not. The probabilty that the first your draw is not faulty is 98/100, not 99/100. Also, because this is sampling without replacement, you now have 97 non-faulty out of 99 total. The probability that the second is also not faulty is 97/99.

To draw "one faulty" you can either draw first a good one and then a bad one or vice versa. The probability that the first you draw is not faulty is, again, 98/100. You now have two faulty ones out of 99 total so the probabilty that the second drawn is faulty is 2/99. Doing it the other way, the probabilty that the first you draw is faulty is 2/100 and then the probability that the second is not is 98/99. Of course, $\displaystyle \frac{98}{100}\frac{2}{99}= \frac{2}{100}\frac{98}{99}$
so the order really doesn't matter. The probability of drawing one faulty item is $\displaystyle 2\left(\frac{98}{100}\frac{2}{99}\right)$.

The probability that the first you draw is faulty is $\displaystyle \frac{2}{100}$ and then the probability that the second is also is $\displaystyle \frac{1}{99}$. The probability that both are faulty is $\displaystyle \frac{2}{100}\frac{1}{99}$.

An electronic components manufacturer submits tenders to three separate companies for a work over the next month. From past experience the contracts manager estimates that the probability that each tender will be accepted as in the table below.
Company Probability
A 0.8
B 0.6
C 0.9

Calculate the probability that at least one tender is accepted:

This is my attempt:
I have assumed this is independent.
Pr(at least one tender) = Pr(A) + Pr(B) + Pr(C) + Pr(A U B) + Pr(A U C) + Pr(B U C) + Pr(A U B U C)

P.S
That doesn't look like a very good method- too complicated.

"At least one" is the same as "NOT none". Calculate the probability that NO tender is accepted (and, of course the probabilities that offers to A, B, and C separately are NOT accepted are 1- .8= .2, 1- .6= .4, and 1- .9= .1, respectively) and subtract from 1. Assuming that these are independent, just multiply the probabilities.

5. ## Re: Calculating probability when sampling without replacement.

Originally Posted by Paymemoney
is that sum P(F = k) calculated as matrices?

because i do not get 1 as the answer
Paymemoney,

Your question makes me suspect that you are not familiar with the notation used by Plato.

In this context,
$\displaystyle \binom{n}{m}$
is not a matrix; it is a binomial coefficient, or equivalently, the number of ways to choose m items out of a set of n objects. So
$\displaystyle \binom{n}{m} = \frac{n!}{m! (n-m)!}$

For example,
$\displaystyle \binom{100}{2} = \frac{100!}{2! 98!} = 4950$

Does that help?

6. ## Re: Calculating probability when sampling without replacement.

thanks for the replys, After i posted my question i worked out the notation that Plato was using