As a whiteboard pen ages, its tip may dry out to the point that the white board pen becomes defective. The company has stock that is 2 years old and, at that age, it is known that 5% of Grade A whiteboard pens will be defective.
A school purchases a box of Grade A whiteboard pens that is 2 yrs old and a class of 26 students is the first to use them.
If every student receives a whiteboard pen from this box, find the probability, that at least one student will receive a defective white board pen?
Now, I know that this is a binomial prob. distribution with n = 26 and p of 0.05, but I can't help oversimplifying the problem by doing 0.05 x 26 which gives me 1 defective pen.
Is the reason that I can't do this is because I am taking a sample and that the sample is not always a representative of the population? Therefore, the prob. of taking a defective pen for each student is always 0.05 hence binomial? And, not that there is 5 % out of the 26 that is defective, right?(as in the number of defective in pop.)
And also question asked for prob. of at least one student therefore 0.05 x 26 wouldn't work as it only gives one answer (if this makes sense).
The problem states "at least one" defective pen. If X is the number of defective pens, the problem is to find $P(X=1) + P(X=2) + ... + P(X=26)$, or more simply, $1.00 - P(X=0)$. $P(X=k)$ is binomial with parameters $n=26$ and $p = 0.05$. $ P(X=0) = $$26 \choose {0}$ $(0.05^0) \times (1.00 - 0.05)^{26}$ and $P(X \neq 0) = 1.00 - P(X = 0) = 1.00 - 0.95^{26}.$