A new surgical procedure is said to be successful 75% of the time. If 4 patients undergo the surgical procedure, what is the probability that

a) all four operations are successful?
b) Exactly three operations are successful?
c) Less than two are successful?


Here $X$ denote the number successful operations out of 4.

Let $p$ be the probability that operation using new surgical procedure is successful.

Given that $p=0.75$ and $n =4$. Thus$X\sim B(4, 0.75)$.

The probability mass function of $X$ is
$$ \begin{aligned} P(X=x) &= \binom{4}{x} (0.75)^x (1-0.75)^{4-x},\\ &\quad x=0,1,\cdots, 4. \end{aligned} $$

a. The probability that all four operations are successful is

$$ \begin{aligned} P(X=4) & =\binom{4}{4} (0.75)^{4} (1-0.75)^{4-4}\\ &=0.3164 \end{aligned} $$

b. The probability that exactly three operations are successful is

$$ \begin{aligned} P(X=3) & =\binom{4}{3} (0.75)^{3} (1-0.75)^{4-3}\\ &=0.4219 \end{aligned} $$

c. The probability that less than two operations are successful is

$$ \begin{aligned} P(X< 2) &=P(X\leq 1)\\ &=P(X=0)+P(X=1)\\ &= 0.0039+0.0469\\ &= 0.0508 \end{aligned} $$

Further Reading