A doctor knows from experience that 10% of of the patients to whom he prescribes a certain blood pressure medication will have undesirable side effects. Calculate the probability that of the 10 randomly selected patients: a) none will have undesirable side effects; b) exactly 4 patients will have undesirable side effects; c) at most 3 will have undesirable side effects; d) at least 3 will have undesirable side effects.

Let the random variable $X$ denote the patients who will undesirable side effects after prescription. $X$ follows a Binomial distribution. That is $X\sim B(n,p)$. For this case, $n=10\\p=0.10$

Therefore, this distribution is given as,

$p(X=x)=\binom{10}{x}0.1^x(1-0.1)^{10-x},\space x=0,1,2,..10$

$a)$

The probability that none will have undesirable side effects is given as,

$p(X=0)=\binom{10}{0}0.1^00.9^{10}=0.9^{10}=0.3487(4dp)$

Therefore, the probability that none of the patients will have undesirable side effects is 0.3487

$b)$

The probability that exactly 4 patients will have undesirable side effects is given as,

$p(X=4)=\binom{10}{4}0.1^40.9^6=210\times0.0001\times0.531441=0.0112(4dp)$

Therefore, the probability that exactly 4 patients will have undesirable side effects is 0.0112.

$c)$

The probability that at most 3 patients will have undesirable side effects is,

$p(X\leqslant3)=\displaystyle\sum^3_{x=0}p(X=x)=\displaystyle\sum^3_{x=0}\binom{10}{x}0.1^x(1-0.1)^{10-x}$

This is equivalent to,

$\binom{10}{0}0.1^0(0.9)^{10}+\binom{10}{1}0.1^1(0.9)^{9}+\binom{10}{2}0.1^2(0.9)^{8}+\binom{10}{3}0.1^3(0.9)^{7}= 0.3486784+ 0.3874205+ 0.1937102+0.05739563=0.9872048$

Therefore, the probability that at most 3 patients will have undesirable side effects is 0.9872048

$d)$

The probability that at least 3 will have undesirable side effects is given as,

$p(X\geqslant3)=\displaystyle\sum^{10}_{x=3}\binom{10}{x}0.1^x(1-0.1)^{10-x}$

$p(X\geqslant3)$ can also be given as,

$p(X\geqslant3)=1-p(X\leqslant2)=1-\{p(X=0)+p(X=1)+p(x=2)\}=1-\{\binom{10}{0}0.1^0(0.9)^{10}+\binom{10}{1}0.1^1(0.9)^{9}+\binom{10}{2}0.1^2(0.9)^{8}\}=1-\{0.3486784+ 0.3874205+ 0.1937102\}=1- 0.9298092=0.0701908$

Therefore, the probability that at least 3 patients will have undesirable side effects is 0.0701908.