Question

suppose 8% of new computers malfunction within the first year. A school buys 15, what's the probability that more than 2 will malfunction in the first year

Answer 1

You need to find the probabilities of 0, 1 and 2 malfunctions. Then these three values of probability are added, and the total subtracted from 1 to find the probability that more than 2 will malfunction in the first year.
Do you want me to give help?

Answer 2

yes please

Answer 3

The probability that a computer will not malfunction is 1.00 - 0.08 = 0.92.
The probability that 0 will malfunction is therefore:
\[P(0\ malfunction)=(0.92)^{15}=you\ can\ calculate\]
Can you do this calculation and post the result?

Answer 4

0.92^15 = 0.28629

Answer 5

Good. Rounded up it is P(0 malfunction) = 0.2863.
Using the binomial distribution the probability that 1 will malfunction is:
\[P(1\ malfunction)=\left(\begin{matrix}15 \\ 1\end{matrix}\right)\times (0.08)^{1} \times (0..92)^{14}=you\ can\ calculate \]

Answer 6

Sorry for the typo. The calculation can be written as:
\[P(1\ malfunction)=15 \times0.08\times(0.92)^{14}=you\ can\ calculate\]

Answer 7

Using the binomial distribution the probability that 2 will malfunction is:
\[P(2\ malfunction)=\left(\begin{matrix}15 \\ 2\end{matrix}\right) \times (0.08)^{2} \times (0.92)^{13}=105\times(0.08)^{2} \times (0.92)^{13}=you\ can\ calculate\]

Answer 8

thank you

Answer 9

You're welcome :)