A coin is tossed three times. Given that at least one head appears, what is the probability that exactly two will appear?

Question

The "at least" confuses me. But I am assuming one head will appear. Making P(first head) = 1.

Correct answer: 3/7

I start with the formula: P(A and B) = P(A) • P(B|A)

Fitting the conditions into this formula, P(A) = 1, P(A and B) I interpret as "exactly two heads".

I found P(A and B) using the complement:

1-3C1(1/2)^3= 5/8 ( 3C1(1/3)^3 represent exactly 1 tail.)

I divided 5/8 by 1, but the answers don't match. ;(

Answer 1

As you said $P(A \cap B)=P(A)\cdot P(B|A)$.

And therefore $P(B|A)=\frac{P(A \cap B)}{P(A)}$, where

$A$: At least one head appears.

$B$: Exactly two head appear.

Applying converse probability.

$P(A)=1-P(\overline A)=\text{1-Probability, that no head appears}=1-0.5^3=0.875$

And $A \cap B$ is the intersection of $A$ and $B$. This is just $B$.

$P(A\cap B)=P(B)= {3 \choose 2}\cdot 0.5^2\cdot 0.5^1=3\cdot 0.5^3=0.375$.

$P(B|A)=\frac{0.375}{0.875}=\frac{3}{7}$

Answer 2

We know that we have at least one $H$, so from the $8$ posible combinations, we remove the one that has $TTT$. From the 7 remaining, we simply count which ones contain two $H$. That would be $HHT$, $HTH$, $THH$.

Alternatively, you could use a random variable $X := \text{number of heads } (\sim Bi(3,\frac{1}{2}))$. Then you look for $P(X = 2 | X > 0)$, which is $\frac{P(X=2)}{P(X>0)}$. If you know how to read tables, then you could simply look up those values.

Hope it helps!