A pharmacist receives a shipment of 23 bottles of a drug and has 3 of the bottles tested. If 6 of the 23 bottles are contaminated, what is the probability that less than 2 of the tested bottles are contaminated? Express your answer as a fraction or a decimal number rounded to four decimal places.

Solution

Given that total number of bottles $m+n = 23$, number of contaminated bottles = $m = 6$ and number of non-contaminated bottles $n = 23 - 6= 17$.

Number of tested bottles $k =3$.

The probability distribution of $X$ (number of contaminated bottles) is Hypergeometric distribution with probability mass function

$$ \begin{aligned} P(X =x) &= \frac{\binom{m}{x}\binom{n}{k-x}}{\binom{m+n}{k}},\\ & & \quad x=0,1,2,3 \end{aligned} $$

The probability that less than 2 of the tested bottles are contaminated is

$$ \begin{aligned} P(X< 2) &= P(X=0) + P(X=1)\\ &=\frac{\binom{6}{0}\binom{17}{3 - 0}}{\binom{23}{3}}+\frac{\binom{6}{1}\binom{17}{3 - 1}}{\binom{23}{3}} \\ &=\frac{\binom{6}{0}\binom{17}{3 }}{\binom{23}{3}}+\frac{\binom{6}{1}\binom{17}{2}}{\binom{23}{3}} \\ &= 0.38396+ 0.46076\\ &= 0.84472 \end{aligned} $$

Thus the probability that less than 2 of the tested bottles are contaminated is $0.84472$.

Further Reading