Probability question?

[JJAduP]

So this probability question was posted in a facebook group and I just can't figure it out. Anyone care to explain how it's solved? I've done some basic probability, so I'm not clueless on the subject, I just have don't this type of problem before.

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 ???

[Umer Bhutto]

if one device has a working to not working ratio of 0.9 : 0.1 Then

10 devices

and if 7 devices have to function and three devices don't function

then

(0.1)^3 * (0.9)^7 = 4.782 * 10^-4

[vekkus4]

It's from the Binomial Distribution

N=10
p = 0.1 = P(doesn't work)
P(X=0) = P(none of them work out of the 10) = 0.1^10
P(X=10) = P(all of them work out of the 10) = 0.9^10
P(X=3) = P(3 of them do not work) = P(7 of them work)
= (ways to choose 3) * (probability the first three don't work) * (probability the last 7 work)
= (0.1)^3 * (0.9)^7 * (10 choose 3)
= 0.05739

10 choose 3 is 120 meaning the pattern of W (working) and N (non-working) devices could be in 120 different patterns.

[Peyush]

probability that device doesn't work properly = 0.1

hence, probability that a device works properly = 1 - 0.1 = 0.9

now, it is given that 7 out of 10 devices function properly but you don't know which 7 devices works properly. So, you have to select randomly 7 devices out of 10 in ^10C_7 ways = 10!/(7! 3!) = 120 ways

required probability = 120 x (0.9)^7 x (0.1)^3 = 0.0574 = 0.057