Question

The probability that an electronic device produced by a company does not function properly is equal to 0.1. If 10 devices are bought, then the probability, to the nearest thousandth, that 7 devices function properly is
A. 0.057
B. 0.478
C. 0.001
D. 0

Answer 1

Answer:

A. 0.057

Step-by-step explanation:

This problem can be result using a Binomial distribution, in which we have n identical events with a probability p of success.

The probability that x of the n events get success is given by:

$P(x)=nCx*p^{x}*(1-p)^{n-x}$

Where nCx can be calculate as:

$nCx=\frac{n!}{x!(n-x)!}$

In this case there a 10 device with a probability 0.9 of function properly and we need to find the probability that 7 of these device function properly, so replacing values, we get:

$P(7)=10C7*0.9^{7}*(1-0.9)^{10-7}$

P(7)=120*0.478*0.001

P(7)=0.057

Finally the probability, to the nearest thousandth, that 7 of the 10 devices function properly is 0.057

Answer 2

The answer is A, 0.057. This one you can solve using logic. D is to low, C is also to low. That leaves A and B. If your rounding to the nearest thousandth C and D don't make to much sense. Looking at it, considering its 7 devices, A makes the most sense. I'll write the math up here later if you want.