HW problem here. I know the answer is 6/16 (per the back of the book) but I can't figure out how they got that.
A fair coin is tossed four times. What is the probability that the
number of heads appearing on the first two tosses is equal to the
number of heads appearing on the second two tosses?
My thought was that the probability of getting x heads on the first two throws would be $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ and the same thing on the other side and then multiply the events together getting $\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$
So where am I going wrong? How do they get the 6?
Thanks.
Best Answer
There are $3$ cases : either you have $0$ head appearing on the first two tosses, either you have $1$ or either you have $2$. You treated the case when you have $2$. The case when you have $0$ gives the same probability of $\frac{1}{16}$. The probability that the number of heads is $1$ on the first two tosses is $2 \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{2}$. Therefore you must add $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{4}{16}$ to $\frac{2}{16}$.