Bulbs

The probability that the bulb can operate 4000 hours is 0.3. What is the probability that exactly one of the eight bulbs can operate 4000 hours?

Correct answer:

p₂ =  19.8 %

Step-by-step explanation:

$$\begin{gathered} p=0.3 \\ {C}_1(8)=\left(\genfrac{}{}{0ex}{}{8}{1}\right)=\frac{8!}{1!(8-1)!}=\frac{8}{1}=8 \\ {p}_2=100\cdot \left(\genfrac{}{}{0ex}{}{8}{1}\right)\cdot p\cdot (1-p{)}^{8-1}=100\cdot 8\cdot 0.3\cdot (1-0.3{)}^{8-1}=19.8\% \end{gathered}$$

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