# Probability (1 Viewer)

#### frog0101

##### New Member
Hi,
A fair coin is tossed n times. What is the probability of heads on the
first toss given that r heads were obtained in the n tosses?

Does anyone know how to do this - solution is r/n

Thanks

#### aa180

##### Member
Hi,
A fair coin is tossed n times. What is the probability of heads on the
first toss given that r heads were obtained in the n tosses?

Does anyone know how to do this - solution is r/n

Thanks
Essentially, the outcome after carrying out the n tosses is a total of r heads (which implies n-r tails). Your task is to determine how many arrangements of the r heads and n-r tails are possible if the first "slot" is taken up by a head, and there is basically no restriction on the remaining r-1 heads and n-r tails. Does that make the question easier for you to answer?

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#### fan96

##### 617 pages
Would it be valid to say that since there were $r$ heads in $n$ tosses, each toss has on average an $r/n$ chance to be heads?

#### aa180

##### Member
Would it be valid to say that since there were $r$ heads in $n$ tosses, each toss has on average an $r/n$ chance to be heads?
If we consider a situation where for every n tosses, we get exactly r heads, then we can calculate

$P(\text{obtaining a head}) = \frac{\text{total number of heads}}{\text{total number of tosses}} = \frac{r}{n}$,

which I believe is exactly what you're saying. So yes, you are right.

The question is worded in a way that suggests the interpretation that we are only considering the cases where we obtain exactly r heads, which is why you can think of it as if that's the only possibility each and every time we toss a coin n times.

#### sida1049

##### Well-Known Member
Would it be valid to say that since there were $r$ heads in $n$ tosses, each toss has on average an $r/n$ chance to be heads?
I'd be very careful about this. Using the term "average" usually implies that you're calculating something via sampling, which is not what you're doing here, since you're after a theoretical answer.

The combinatorial reasoning aa180 provided is a good one, as it leads you unequivocally to the answer via the computation
$\frac{\binom{n-1}{r-1}}{\binom{n}{r}} = \frac{r}{n}.$