# Thread: 3 of 10 samples are contaminated

1. ## 3 of 10 samples are contaminated

3 of 10 samples are contaminated. 2 are randomly chosen. What is the probability of both of the chosen samples are contaminated?

Hope you can help.
Thanks.

EDITlease include all working.

2. Originally Posted by Bartimaeus
3 of 10 samples are contaminated. 2 are randomly chosen. What is the probability of both of the chosen samples are contaminated?

Hope you can help.
Thanks.
Probability that first sample is contaminated P1=3/10.

Probability that second sample is contaminate given the first
is contaminated is P2=2/9.

Probability both contaminated=P1*P2

RonL

3. Thanks RonL

4. Hello, Bartimaeus!

RonL's solution is the best one.
Here's an alternate approach (if you're familiar with Combinations).

Three of ten samples are contaminated. .Two are randomly chosen.
What is the probability of both of the chosen samples are contaminated?

There are $\displaystyle \binom{10}{2} = 45$ possible outcomes.

To get two contaminated samples, there are: $\displaystyle \binom{3}{2} = 3$ ways.

Therefore: .$\displaystyle P(\text{both contaminated}) \:=\:\frac{3}{45} \:=\:\frac{1}{15}$

5. I'm not really familiar with Combinations.
Could you explain them further to me, please?

6. Originally Posted by Bartimaeus
I'm not really familiar with Combinations.
Could you explain them further to me, please?
Say there are 5 bullets:
.36 .357 .308 .45 .50
And you are told to select two of them, the possibilities are,
Code:
.36 and .357
.36 and .308
.36 and .45
.36 and .50
.357 and .308
.357 and .45
.357 and .50
.308 and .45
.308 and .50
.45 and .50
In total there are 10 possibilities.
You chose 2 from 5. We write,
$\displaystyle {5 \choose 2}=10$ another notation,
$\displaystyle _5C_2=10$.

The mathematical formula is,
$\displaystyle {n\choose m}=\frac{n!}{m!(n-m)!}$

Note: Sometimes the combinations formula is refferred to as the binomial coefficients. Because the coefficients in the binomial expansion follow the combinations formula.