1. ## Probability given that...

Here's the question. Ive been tryin to solve it but to no avail. How toyou approach it?

A two-sided coin is flipped four times. Given that the coin landed heads up more than twice, what is the probability that it landed heads up all four times?
Thanx

2. If we are given that two of the coins landed head up, in order that all four land head up we only need that the other two coins landed head up. Since the flips are independent, that is exactly the same as the probability that, when flipping 2 coins, both land head up. what is that?

3. Originally Posted by raxrox
Here's the question. Ive been tryin to solve it but to no avail. How toyou approach it?

A two-sided coin is flipped four times. Given that the coin landed heads up more than twice, what is the probability that it landed heads up all four times?
Thanx
Let X be the random variable number of heads.

X ~ Binomial(n = 4, p = 1/2)

You need to calculate $\Pr(X = 4 | X \geq 3)$.

$\Pr(X = 4 | X \geq 3) = \frac{\Pr(X = 4 ~ \text{and} ~ X \geq 3)}{\Pr(X \geq 3)} = \frac{\Pr(X = 4)}{\Pr(X \geq 3)} = \, ....$

4. Originally Posted by HallsofIvy
If we are given that two of the coins landed head up, in order that all four land head up we only need that the other two coins landed head up. Since the flips are independent, that is exactly the same as the probability that, when flipping 2 coins, both land head up. what is that?
Not so, unfortunately. We're given that more than two of the coins landed heads up.

5. There is very simple way to view this problem.
To have more than two heads: HHHT, HHTH, HTHH, THHH, or HHHH.
Now what is the probability?