Question

A new surgery is successful 75% of the time. if the results of 10 such surgeries are randomly sampled, what is the probability that fewer than 9 of them are successful

Answer 1

$\text{Binomial Distribution: } P(X = r) = \binom{n}{r}(p)^{r}(1 - p)^{n - r}$
$P(X < 9) = 1 - P(X = 9) - P(x = 10)$

$P(X = 9) = \binom{10}{9}(0.75)^{9}(0.25)$
$P(X = 10) = \binom{10}{10}(0.75)^{10}$

$P(X < 9) = 1 - \binom{10}{9}(0.75)^{9}(0.25) - \binom{10}{10}(0.75)^{10}$
$\approx 0.755974769...$