If a coin is tossed four times, what is the probability that heads will come up twice?
I cant quite seem to get the right answer. anything will help
This is a binomial probability.
The prob of getting exactly two heads is
$\displaystyle \left( \frac{1}{2}\right)^2 \left( \frac{1}{2}\right)^2$
but this $\displaystyle \binom{4}{2}=6$ ways
so we get
$\displaystyle 6 \cdot \left( \frac{1}{2}\right)^4=\frac{6}{16}=\frac{3}{8}$
$\displaystyle \Omega = $ HHHH ; HHHT ; HHTH ; HTHH ; HHTT ; HTTT ; HTTH ; HTHT ; TTTT ; TTTH ; TTHT ; THTT ; TTHH ; THHH ; THHT ; THTH
Let A be the event that heads come up twice.
$\displaystyle A =$ HHTT ; HTTH ; HTHT ; TTHH ; THHT ; THTH
$\displaystyle P(A) = \frac{6}{16} = \frac{3}{8}$