# Products from a certain machine are too large 15% of the time. What is the probability that in a run of 20 parts, 5 are too large?

${C}_{5}^{20} {\left(.15\right)}^{5} {\left(.85\right)}^{15} \approx .103$
The probability that 5 parts are too large is $.15 \times .15 \times .15 \times .15 \times .15 = {\left(.15\right)}^{5}$ and since 5 parts are too large, the other 15 are of acceptable size with probability ${\left(1 - .15\right)}^{15}$. Since any 5 parts can be too large in no particular order, there are ${C}_{5}^{20}$ ways to do this, so the probability that exactly 5 parts are too large is ${C}_{5}^{20} {\left(.15\right)}^{5} {\left(.85\right)}^{15} \approx .103$.