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

2. Originally Posted by rkaminky
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

$\left( \frac{1}{2}\right)^2 \left( \frac{1}{2}\right)^2$

but this $\binom{4}{2}=6$ ways

so we get

$6 \cdot \left( \frac{1}{2}\right)^4=\frac{6}{16}=\frac{3}{8}$

3. The above post if for exactly two heads. If you want 2 or more you need to add the prob of 3 heads and prob of 4 heads.

Good luck.

4. Originally Posted by rkaminky
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
Many approaches are possible. Are you familiar with the binomial distribution?

Alternatively, consider all the different arrangements of HHTT. There are six. The probability of each is (1/2)^4 (assuming a fair coin). Therefore .....

5. Originally Posted by rkaminky
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
$\Omega =$ 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.

$A =$ HHTT ; HTTH ; HTHT ; TTHH ; THHT ; THTH

$P(A) = \frac{6}{16} = \frac{3}{8}$

6. Originally Posted by mr fantastic
Many approaches are possible. Are you familiar with the binomial distribution?
Let us try Mr F's suggestion of binomial distribution.

$P(T) = {4 \choose 2} \left( \frac{1}{2} \right) ^{2} \left( \frac{1}{2} \right) ^{2} = \frac{6}{16} = \frac{3}{8}$