Probability of flipping a coin an exact number of times A biased coin flips heads with probability 4/9 and tails with probability 5/9. The coin is flipped 90 times. What is the probability that heads is flipped exactly 40 times?
I believe this question requires the utilisation of the binomial distribution in which:
$Pr(H) = \frac{4}{9}$
$Pr(T) = \frac{5}{9}$
$n = 90$
$k = 40$
$Pr(X = 40) = {90\choose 40}(\frac{4}{9})^{40}  (1 - \frac{4}{9}) ^{90-40} =
{90\choose 40}  (\frac{4}{9})^{40} (\frac{5}{9}) ^{50}$
is approximately 0.084
Am I going abouts this correctly? Not feeling too confident

The standard deviation for this is $$\sqrt{np(1-p)} = \sqrt{90\cdot \frac{4}{9}\cdot\frac{5}{9}} = 4.714$$
Roughly $2/3$ of the outcomes for the number of heads would be between about $35$ and $45$. That's $11$ outcomes each with a probability of less than $0.09$ to ensure the total is less than $1$. So your answer of $.084$ is right in the ball park.
