# random variable of pens

There are 20 pens in a box and 3 of them are without ink.

1.If I pick 3 pens randomly, what’s the probability that all the pens can write.

2.I pick a pen randomly and then I replace it back to the box. I repeat this procedure six times.calculate the probability that we picked at most one pen that with no ink

3.use a Poisson random variable to calculate it again

4.If I pick pens and replace them, how many pens do I have to pick on average until picking up the third pens without ink?

5.Pick out three pens together, and put them back to the basket if not all three are without ink. How many times will you repeat this procedure on average?

I think question one is a hypergeometric distribution

question 2 is a binominal distribution and what should I use in passion random variable since variance is not equal to mean and I don't know what should I do for the question 4 and 5

• For the first : three from seventeen. Commented Oct 30, 2016 at 23:13
• @AbdallahHammam: Not an answer, possibly a cryptic partial clue? Commented Oct 30, 2016 at 23:53

For 3 use $\lambda=np$, fourth is negative binomial and fifth is geometric, i think.

• Simultaneous with mine (+1). Commented Oct 30, 2016 at 23:55

Here are some answers and clues:

(1) ${17 \choose 3}{3 \choose 0}/{20 \choose 3} \approx 0.60.$ Suggest you compute it to four places. Right $X$ is hypergeometric and you want $P(X = 3).$

(2) $X$ is the number of pens chosen that have no ink. $X \sim Binom(6, 3/20).$ Then $P(X \le 1) = P(X=0) + P(X=1) = 0.7765.$

(3) "It" is not perfectly clear. If it means to use a Poisson approximation to (2), then $Y \sim Pois(\lambda = 18/20)$ and $P(Y\le 1) = 0.7725.$

(4) Negative binomial distribution is correct. Use $p = P(\text{No ink}) = 3/20$ and you're waiting for $r = 3$ no-ink pens.

(5) "Repeat" in order to get what? Possibly geometric with $1-p$, where $p$ is the answer in (1), and waiting for first draw with some working pens.

• for the question 5 it means how many time I need to do the picking so that I can pick out three pens that without ink Commented Oct 31, 2016 at 7:19
• Thanks. That is what I guessed -- and said in (5) after "Possibly ..." Also, I suppose what @Momo assumed. Commented Oct 31, 2016 at 7:46
• but I was wondering how to classify the question to different kind of distribution Commented Oct 31, 2016 at 7:48
• The point of the overall question is to provide drill on various kinds of distributions. Part (5) is geometric (a special case of negative binomial). Commented Oct 31, 2016 at 8:07
• sorry that is the fourth question asking for the expected value of negative binomial Commented Oct 31, 2016 at 16:58