A notepad manufacturer requires that 90% of the production is of sufficient quality. To check this, 12 computers are chosen at random every day and tested thoroughly. The day's production is deemed acceptable if at most 1 computer fails the test. If more computers fail, all the computers produced during the day need to be tested.
a) What is the probability that a single day production passes the quality test if only 80% of the computers are up to standards?
b) What is the probability that all the production from a single day has to be tested if 90% of the computers are up to standards?
This is a bit confusing.
My attempt to A. There are twelve trials (n=12), 80% of tested computers is 9.6 (round to 10), which means that k = 10 and n-k = 2, p = 0.8 and q = 1-p = 0.2. Therefore C(12,10)*(0.8)^10 * (0,2)^2 = 0.07 = 7%. I am not sure if I am thinking correctly about it. Can you help me guys? Or at least give me any kind of hint. I would really appreciate that!
$a)$ the probability that a computer is correct is $0.8$, hence the products of that day will pass the quality test if at most one of the computers is out of service. Hence:
$Pr(passing\space the\space quality\space test)=(0.8)^{12}+C(12,1)(0.8)^{11}(0.2)$, in which $C(12,1)=\frac{12!}{1!\times (12-1)!}$
$b)$ In this case, the first testing among 12 computers, must be failed. Hence:
$Pr(not\space passing\space the \space quality\space test)=1-Pr(passing\space the \space quality\space test)=1-(0.9)^{12}-C(12,1)(0.9)^{11}(0.1)$