Respuesta :
To solve the exercise, you can use the formula of the binomial distribution:
[tex]\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}[/tex]So, in this case, we have:
[tex]\begin{gathered} n=10 \\ p=\frac{6}{10}=0.6 \end{gathered}[/tex]Because "success" is that there are defective cameras, 6 defective cameras out of 10 in total.
For part A, we have:
[tex]\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}[/tex]Therefore, the probability of getting 2 defective cameras is 0.0106.
For part B, we have:
[tex]\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}[/tex]Therefore, the probability of getting one defective camera is 0.0016.
