Hospital records show that of patients suffering from a certain disease, \displaystyle{75}\%75% die of it. What is the probability that of \displaystyle{6}6 randomly selected patients, \displaystyle{4}4 will recover?

 Hospital records show that of patients suffering from a certain disease, 

$\displaystyle{75}\%$ die of it. What is the probability that of $\displaystyle{6}$ randomly selected patients, $\displaystyle{4}$ will recover?

Solution

This is a binomial distribution because there are only 2 outcomes (the patient dies, or does not).

Let X = number who recover.

Here, $\displaystyle{n}={6}$ and $\displaystyle{x}={4}$. Let $\displaystyle{p}={0.25}$ (success, that is, they live), $\displaystyle{q}={0.75}$ (failure, i.e. they die).

The probability that $\displaystyle{4}$ will recover:

$\displaystyle{P}{\left({X}\right)}$ $\displaystyle={{C}_{{x}}^{{n}}}{p}^{x}{q}^{{{n}-{x}}}$ $\displaystyle={{C}_{{4}}^{{6}}}{\left({0.25}\right)}^{4}{\left({0.75}\right)}^{2}$ $\displaystyle={15}\times{2.1973}\times{10}^{ -{{3}}}$ $\displaystyle={0.0329595}$

Histogram of this distribution:

We could calculate all the probabilities involved and we would get:

$\displaystyle{X}$	$\displaystyle\text{Probability}$
$\displaystyle{0}$	$\displaystyle{0.17798}$
$\displaystyle{1}$	$\displaystyle{0.35596}$
$\displaystyle{2}$	$\displaystyle{0.29663}$
$\displaystyle{3}$	$\displaystyle{0.13184}$
$\displaystyle{4}$	$\displaystyle{3.2959}\times{10}^{ -{{2}}}$
$\displaystyle{5}$	$\displaystyle{4.3945}\times{10}^{ -{{3}}}$
$\displaystyle{6}$	$\displaystyle{2.4414}\times{10}^{ -{{4}}}$

The histogram is as follows:

Open image in a new page

Histogram of the binomial distribution

It means that out of the $\displaystyle{6}$ patients chosen, the probability that:

• None of them will recover is $\displaystyle{0.17798}$,
• One will recover is $\displaystyle{0.35596}$, and
• All $\displaystyle{6}$ will recover is extremely small.