Question

(A) A shipment of 10 cameras will likely have 6 defectives. If a person buys 2 cameras, what is the probability of getting 2 defectives?(B) What are the odds in favor of getting a defective camera?

Answer 1

To solve the exercise, you can use the formula of the binomial distribution:

`\begin{gathered} P(x)=\binom{n}{x}p^x(1-p)^(n-x) \\ \text{ Where } \\ n\text{ is the number of trials (or the number being sampled)} \\ x\text{ is the number of successes desired} \\ p\text{ is the number of getting a success in one trial} \end{gathered}`

So, in this case, we have:

`\begin{gathered} n=10 \\ p=(6)/(10)=0.6 \end{gathered}`

Because "success" is that there are defective cameras, 6 defective cameras out of 10 in total.

For part A, we have:

`\begin{gathered} x=2 \\ P(x)=\binom{n}{x}p^x(1-p)^(n-x) \\ P(2)=\binom{10}{2}\cdot0.6^2\cdot(1-0.6)^(10-2) \\ P(2)=\binom{10}{2}\cdot0.6^2\cdot(0.4)^8 \\ P(2)=45\cdot0.36\cdot0.00065536 \\ P(2)=0.0106 \end{gathered}`

Therefore, the probability of getting 2 defective cameras is 0.0106.

For part B, we have:

`\begin{gathered} x=1 \\ P(x)=\binom{n}{x}p^x(1-p)^(n-x) \\ P(1)=\binom{10}{1}\cdot0.6^1\cdot(1-0.6)^(10-1) \\ P(1)=\binom{10}{1}\cdot0.6^1\cdot(0.4)^9 \\ P(1)=10\cdot0.6\cdot0.000262144 \\ P(1)=0.0016 \end{gathered}`

Therefore, the probability of getting one defective camera is 0.0016.