# Thread: Probability question involving number of permutations

1. ## Probability question involving number of permutations

Hi

my question is From a group of 12 children, two random samples of size three are chosen; the first sample being replaced before the second sample is chosen.

a) Four of the group are named Matthew, Mark, Luke and John. Calculate the probability that none of these boys appear in the first sample.
[6 marks]

b) Calculate the probability that the samples have at least one person in common.
[9 marks]

for part a) i know that Number of ways of choosing a sample of 3 is $\displaystyle \dbinom{12}{3}=220$

but i am not sure what to do next and i don't know how to do part b)

thanks

2. Originally Posted by rpatel
From a group of 12 children, two random samples of size three are chosen; the first sample being replaced before the second sample is chosen.

a) Four of the group are named Matthew, Mark, Luke and John. Calculate the probability that none of these boys appear in the first sample.
The answer to part a) is $\displaystyle \frac{\binom{8}{3}}{\binom{12}{3}}$.
Now to get more help, reply with reasons that answer is correct.

3. For part a)

there are 8 remaining children.

there are $\displaystyle \binom{8}{3}$ ways to choose 3 from the remaining 8.

Hence the probability is $\displaystyle \frac{\binom{8}{3}}{\binom{12}{3}}$

The chance of both groups not containing a common person is

$\displaystyle \frac{\binom{12}{3}\binom{9}{3}}{\binom{12}{3}^2}$

since $\displaystyle \binom{12}{3}$ is the number of ways of choosing 3 from 12,

$\displaystyle \binom{9}{3}$ is the number of ways of choosing 3 from the non-matching 9.
These can be paired up so we multiply to find the total number of combinations.

In this case, the denominator consists of pairing any 3 with any 3.

Now, you should be able to answer b)

4. so will the answer be $\displaystyle 1-\frac{\binom{12}{3}\binom{9}{3}}{\binom{12}{3}^2}$ ?

5. I would prefer you try the calculations as Plato outlined,
The answer is not so important as getting there.
Getting through working out the logic for this example
will stand to you as you apply yourself to other examples later.

Why are the denominators so?

6. For part a) my understanding of it is there was 4 people out 12 which are not chosen in the first sample$\displaystyle \dbinom{4}{0}$ . Hence this must mean that from the remaining 8 the the sample of 3 must be chosen$\displaystyle \dbinom{8}{3}$. Hence this means $\displaystyle \frac{\dbinom{4}{0}\dbinom{8}{3}}{\dbinom{12}{3}}$ which gives $\displaystyle \frac{14}{25}$.

However i need help with the understanding of part b)

7. b) Calculate the probability that the samples have at least one person in common.

For part b, take the complement event
so from my understanding,
it would be 1 - (probability of none in common).

8. Yes,
differentiate is correct as taking the complement event is often easier
to calculate.

You're showing the right spirit now rpatel,
however you need to calculate the numerator and denominator more carefully.

Try it again and we will soon get through part b).
Also, that's correct,
the denominator is the number of ways of selecting 3 from the complete 12.
Your calculations for 3 from 8 and 3 from 12 need to be reviewed.

9. $\displaystyle \frac{\binom{8}{3}}{\binom{12}{3}}=\frac{(\frac{8! }{5!3!})}{(\frac{12!}{9!3!})}=\frac{8(7)6}{12(11)1 0}$

10. Ok yep i the answer should have been $\displaystyle \frac{14}{55}$, i made a calculator error.

Ok here is my understanding of part b) In the first sample 3 are chosen from 12 hence $\displaystyle \dbinom{12}{3}$. Then in the second sample a different 3 are chosen from remaining 9 students hence $\displaystyle \dbinom{9}{3}$ therefore the probability of the samples not containing a common person is $\displaystyle \frac{\binom{12}{3}}{\binom{12}{3}}\times\frac{\bi nom{9}{3}}{\binom{12}{3}}$

Hence the Probability of the sample containg at least one person in common is $\displaystyle 1-\frac{\binom{12}{3}}{\binom{12}{3}}\times\frac{\bi nom{9}{3}}{\binom{12}{3}}$

is my explanation correct ?

11. Good!

Now you can see the redundancy of the left side.
So, these are not two probabilities multiplied together
as is most often the case.

Any of the first selection of 3 from 12 can be paired with any selection of
3 from the other 9 since those 9 are all different.

That gives us the number of selections of groups of distinct 6's.

The number of total possible groups of 6's goes in the denominator position,
this includes all groups of 6 with replication.
This is all selections of 3 from 12 paired with all selections of 3 from 12.
This is because the first 3 are replaced.