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.
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.
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: .$\displaystyle P(\text{both contaminated}) \:=\:\frac{3}{45} \:=\:\frac{1}{15}$
Say there are 5 bullets:Originally Posted by Bartimaeus
.36 .357 .308 .45 .50
And you are told to select two of them, the possibilities are,
In total there are 10 possibilities.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
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.