since 6% of the alarms are defective, probability of failure =q=0.06
p = 0.94
k =number of successes in n trials= 24-2=22
Use the pmf of the binomial distribution to find P(K=22)
A manufacturer produces 24 yard alarms per week. Six percent of all the alarms produced are defective. What is the probability of getting two defective alarms in one week?
I've identified the parameters
n = number of trials = 24
p = the probability of success p = 0.06
q=1-p = probability of failure = 0.94
x = the number of sccesses in n trials and 0<=x <= n x=2
P(x) probability of getting exatly x successes in n trials(I have to find out)
I can calculate P(x) using IT(Winstats) but I don't have it, any other ways to calculate it?