Coint toss probability question

[fraggle]

Homework Statement

A coin is tossed 3 times. at least 1 head is obtained. Determine the probability that exactly 1 head is obtained

Homework Equations

The Attempt at a Solution

brute force indicates that there 7 possible combinations:

HHH, HHT, HTH, HTT, THH, THT, TTH (because at least one heads is obtained).

Out of these we see that there are 3 occasions where there is exactly one heads.
If we let S be the sample space of all possable outcomes, and let E be the event of having exactly one heads, then a basic rule of probability indicates that the probability P(E)=|E|/|S|=3/7.

However, using another method:
fix one coin throw's result as heads. Then the probability of the other two throws being tails is (1/2)(1/2)=1/4.
Since we can fix heads 3 times, this indicates that the probability is 3(1/4)=3/4.

Can anyone spot where the flaw is in either of these attempts at a solution?
thanks
(this isn't a homework question btw)

 

[housemartin]

3/7 is when you drop "at least 1 head is obtained" out
1/4 is when when "at least 1 head is obtained" and that is the only head obtained

 

[D H]

However, using another method:
fix one coin throw's result as heads. Then the probability of the other two throws being tails is (1/2)(1/2)=1/4.
Since we can fix heads 3 times, this indicates that the probability is 3(1/4)=3/4.

You can't do that. Probabilities add only if the underlying events are mutually exclusive events in the same parent population. That isn't the case here.

 

[fraggle]

3/7 is when you drop "at least 1 head is obtained" out
1/4 is when when "at least 1 head is obtained" and that is the only head obtained

Can you please explain?

 

[hellofolks]

I think the flaw (in the second attempt) is the following:

You summed the probabilities of events {HTT}, {THT} and {TTH} because they are disjoint events whose union is the event targeted. The error was to consider each of them as intersections of two equally likely, independent events whose probability is 1/2 (getting two tails).
Rather, they are individual instances of a sample space with equally likely 7 elements as you deduced directly. So their probabilities are 1/7 and the desired probability equals 3(1/7)=3/7.

Have you learned random variables? If you have, I guess there's an easier way to do this exercise. Let X denote the number of heads in the three throws. Then, as each of the 3 throws is independent and the probability of obtaining a head in each of them remains constant and equal to 1/2, X is Binomial with parameters n=3 and p=1/2. The desired probability is then

P(X=1|X\geq 1)=\frac{P(X=1,X\geq 1)}{P(X\geq 1)}=\frac{P(X=1)}{1-P(X=0)}.

But since P(X=k)={n \choose k}p^k(1-p)^{n-k} for a Binomial variable, we have P(X=0)=1/8 and P(X=1)=3/8. So,

P(X=1|X\geq 1)=\frac{3/8}{1-1/8}=\frac{3/8}{7/8}=\frac{3}{7}.

Notice the comma in P(X=1,X\geq 1) means intersection.

 

[fraggle]

I think the flaw (in the second attempt) is the following:

You summed the probabilities of events {HTT}, {THT} and {TTH} because they are disjoint events whose union is the event targeted. The error was to consider each of them as intersections of two equally likely, independent events whose probability is 1/2 (getting two tails).
Rather, they are individual instances of a sample space with equally likely 7 elements as you deduced directly. So their probabilities are 1/7 and the desired probability equals 3(1/7)=3/7.

Have you learned random variables? If you have, I guess there's an easier way to do this exercise. Let X denote the number of heads in the three throws. Then, as each of the 3 throws is independent and the probability of obtaining a head in each of them remains constant and equal to 1/2, X is Binomial with parameters n=3 and p=1/2. The desired probability is then

P(X=1|X\geq 1)=\frac{P(X=1,X\geq 1)}{P(X\geq 1)}=\frac{P(X=1)}{1-P(X=0)}.

But since P(X=k)={n \choose k}p^k(1-p)^{n-k} for a Binomial variable, we have P(X=0)=1/8 and P(X=1)=3/8. So,

P(X=1|X\geq 1)=\frac{3/8}{1-1/8}=\frac{3/8}{7/8}=\frac{3}{7}.

Notice the comma in P(X=1,X\geq 1) means intersection.

Thanks a lot, that made sense.