1. ## P(1 Head, 2 Tails)

A fair coin is tossed three times. What is the probability of obtaining one Head and two Tails? (A fair coin is one that is not loaded, so there is an equal chance of it landing Heads up or Tails up.) Also, is this situation considered to be independent events?

Solution:

I found the formula for the probability of A and B online and it is P(A and B) = P(A) * P(B).

1. I think this situation is considered independent events. It is the probability of Event A times the probability of Event B.

2. See below.

HHH
HHT
HTH
TTT
THH
THT
TTH
HTT

P(1 H & 2 Tails) = 3/8.

Yes?

Is there a way to solve this question using a probability formula?

2. ## Re: P(1 Head, 2 Tails)

Originally Posted by harpazo
A fair coin is tossed three times. What is the probability of obtaining one Head and two Tails? (A fair coin is one that is not loaded, so there is an equal chance of it landing Heads up or Tails up.) Also, is this situation considered to be independent events?
Solution:
I found the formula for the probability of A and B online and it is P(A and B) = P(A) * P(B).
1. I think this situation is considered independent events. It is the probability of Event A times the probability of Event B.
2. See below.
HHH
HHT
HTH
TTT
THH
THT
TTH
HTT
P(1 H & 2 Tails) = 3/8.
Yes? Is there a way to solve this question using a probability formula?
If you are going to work these you must learn counting theory.
How many ways are there to arrange the strung $HTT$?
Well $\dfrac{3!}{2!\cdot 1!}=3$ out of $2^3=8$
So yes you are correct.
If it were six heads & four tails what is the answer?

3. ## Re: P(1 Head, 2 Tails)

The probability of "one head, two tails" in that order is (1/2)(1/2)(1/2)= 1/8. It is not difficult to see that that is the probability for any specific order. How many orders are there?
It is easy to write them: HTT, THT, TTH. Three possible orders. That is the same as $\displaystyle \begin{pmatrix}3 \\ 2\end{pmatrix}= \begin{pmatrix} 3\\ 1\end{pmatrix}= \frac{3!}{(1!)(2!)}= 3$.

So the probability of "one head, two tails" is $\displaystyle \frac{3}{8}$. In general if you flip n times, the probability or 'h heads, t tails", where h+ t= n, is $\displaystyle \frac{\begin{pmatrix}n \\ h \end{pmatrix}}{2^n}= \frac{n!}{h!t!2^n}$.

Even more generally if you do something that has probability p of outcome A and probability q of outcome B, where p+ q= 1, N times, the probability of outcome A happening n times and outcome B happening m times, with m+ n= N, is $\displaystyle \begin{pmatrix} N \\ n\end{pmatrix}p^nq^m= \frac{N!}{n!m!}p^nq^m$.

4. ## Re: P(1 Head, 2 Tails)

Originally Posted by HallsofIvy
The probability of "one head, two tails" in that order is (1/2)(1/2)(1/2)= 1/8. It is not difficult to see that that is the probability for any specific order. How many orders are there?
It is easy to write them: HTT, THT, TTH. Three possible orders. That is the same as $\displaystyle \begin{pmatrix}3 \\ 2\end{pmatrix}= \begin{pmatrix} 3\\ 1\end{pmatrix}= \frac{3!}{(1!)(2!)}= 3$.

So the probability of "one head, two tails" is $\displaystyle \frac{3}{8}$. In general if you flip n times, the probability or 'h heads, t tails", where h+ t= n, is $\displaystyle \frac{\begin{pmatrix}n \\ h \end{pmatrix}}{2^n}= \frac{n!}{h!t!2^n}$.

Even more generally if you do something that has probability p of outcome A and probability q of outcome B, where p+ q= 1, N times, the probability of outcome A happening n times and outcome B happening m times, with m+ n= N, is $\displaystyle \begin{pmatrix} N \\ n\end{pmatrix}p^nq^m= \frac{N!}{n!m!}p^nq^m$.
This is a bit too mathematical.

5. ## Re: P(1 Head, 2 Tails)

Originally Posted by harpazo
This is a bit too mathematical.
Harpy, stick to basic algebra for now...

6. ## Re: P(1 Head, 2 Tails)

Okay, from now on, I will give geological answers!

7. ## Re: P(1 Head, 2 Tails)

Originally Posted by DenisB
Harpy, stick to basic algebra for now...
No need to be rude.