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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

The 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

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

$X \sim B(100, 0.5)$

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

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

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

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

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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).

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  • $\begingroup$ thanx david ! $\endgroup$ – Zingo Jul 17 '15 at 7:08
  • $\begingroup$ there is a minor rounding issue: $\frac{0.5397946}{0.9823999}\approx 0.5494653$ would be closer $\endgroup$ – Henry Jan 10 at 21:18
  • $\begingroup$ Thanks for that @Henry. Of course the probability will actually be a rational number. Do you think you could calculate that and post it, so that we have no error at all? $\endgroup$ – David Jan 13 at 23:02

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