The Question:

You flip a fair coin $100$ times. What is the probability of getting at least $50$ heads given that you have at least $40$ heads.

Hi .. Im new to this world called probability. I'm trying to solve this question as binomial distribution . yet Im not sure if my thoughts are right, hope you guys check my answer

sample of $n = 100$ of independent trials each of which can have only two possible outcomes, which are either “head” or “tail” >> its binomial distribution

$n = 100,p = 0.5,q = 0.5$

$X \sim B(100, 0.5)$

$P(A) = P(x\ge50) = 1 - P(x\le49) = 1 – .460 = .540$

$P(B) = P(x\ge40) = 1- P(x\le39) = 1 – 0.018 = 0.982$

$P(A \cap B) = P( 50\ge x\ge40) = P(x\le50) - P(x\le39) = 0.540 – 0.018 = 0.522$

$P(A|B) = P(A \cap B) / P(B) = 0.522 / 0.982 = 0.531$

The event $A\cap B$ is the event that the number of heads is at least $50$, and at least $40$. This is the same as saying that the number is at least $50$. That is, $A\cap B=A$ and so $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}=\frac{0.540}{0.982}=0.550$$ (assuming your calculations for $P(A)$ and $P(B)$ are correct - I didn't check them).
• there is a minor rounding issue: $\frac{0.5397946}{0.9823999}\approx 0.5494653$ would be closer – Henry Jan 10 at 21:18