Question

Ten artists (of two different experience levels) are exhibiting their paintings in an art gallery. Three have exhibited in previous shows (making them “experienced artists”) and seven have not. If a patron selects three paintings to purchase at random, find the probability of selecting:

(A) the probability of selecting at least 2 experienced artists’ paintings.

(B) the probability of selecting no experienced artists’ paintings.

(C) the probability of selecting some experienced artists’ paintings.

(D) the probability of selecting less than 2 experienced artists’ paintings.

I also have an additional question, is this a geometric, hypergeometric, Poisson, normal, or a standard normal problem? I'm very confused on how to tell the difference between them. Thank you very much. Also if it appropriate to the question. What would be the success or failure part of this problem, described in words.

Answer 1

There are a total of 10 artists of which 3 are experienced artists and 7 are not experienced.

The patron selects 3 paintings. Let X be the number of experienced artists' painting selected (out of 3). Here, a success would be the event of selecting a painting of experienced artist and the failure would be not selecting a painting of experienced artist

Number of ways of selecting 3 paintings out of 10 is

10 3

Number of ways of ways of selecting X=x experienced artists' painting (and 3-x paintings from others) is

$\binom{3}{x}\binom{7}{3-x}$

The probability of selecting X=x experienced artists' painting is

$\begin{align*} P(X=x)&=\frac{\text{Number of ways of ways of selecting X=x experienced artists' painting}}{\text{total number of ways of selecting 3 paintings}}\\ &=\frac{\binom{3}{x}\binom{7}{3-x}}{\binom{10}{3}},\quad x=0,1,2,3 \end{align*}$

This is the formula for Hypergeometric distribution.

(A) the probability of selecting at least 2 (which is 2 or more) experienced artists’ paintings.

$\begin{align*} P(X\ge 2)&=P(X=2)+P(X=3)\quad \text{x can take 0,1,2,3}\\ &=\frac{\binom{3}{2}\binom{7}{3-2}}{\binom{10}{3}}+\frac{\binom{3}{3}\binom{7}{3-3}}{\binom{10}{3}}\\ &=\frac{\frac{3!}{2!(3-2)!}\times \frac{7!}{1!(7-1)!}}{\frac{10!}{3!(10-3)!}}+\frac{\frac{3!}{3!(3-3)!}\times \frac{7!}{0!(7-0)!}}{\frac{10!}{3!(10-3)!}}\\ &=0.175+0.0083\\ &=0.1833 \end{align*}$

ans: the probability of selecting at least 2 experienced artists’ paintings is 0.1833

(B) the probability of selecting no experienced artists’ paintings.

$\begin{align*} P(X=0)&=\frac{\binom{3}{0}\binom{7}{3-0}}{\binom{10}{3}}\\ &=\frac{\frac{3!}{0!(3-0)!}\times \frac{7!}{3!(7-3)!}}{\frac{10!}{3!(10-3)!}}\\ &=0.2917 \end{align*}$

ans: the probability of selecting no experienced artists’ paintings is 0.2917

(C) the probability of selecting some experienced artists’ paintings is same as the probability of selecting 1 or more experienced artists’ paintings

$\begin{align*} P(X\ge 1)&=1-P(X=0)\\ &=1-0.2917\quad\text{from part b)}\\ &=0.7083 \end{align*}$

ans: the probability of selecting some experienced artists’ paintings is 0.7083

(D) the probability of selecting less than 2 (does not include 2) experienced artists’ paintings

$\begin{align*} P(X<2)&=1-P(X\ge 2)\\ &=1-0.1833\quad \text{ from part a)}\\ &=0.8167 \end{align*}$

ans: the probability of selecting less than 2 experienced artists’ paintings is 0.8167

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What type of distribution X follows?

The problem of having X number of successes in n trials is Binomial only when the success probability remains the same and the event has just 2 outcomes, success, failure. Here, the success is, selecting experienced artists' painting. But can we say the probability remains the same?

When we select the first painting, the probability of selecting experienced artists' painting is 3/10=0.3

When we select the second painting, the probability depends on what we have done in the first. If the first painting selected is of an experienced artist, then we are left with 2 paintings by experienced artists and hence the probability of selecting  experienced artists' painting in the second round given that in the first round we selected an experienced artists' painting is 2/9

So we can see that the probability of selecting experienced artists' painting changes in each round and is dependent on what has been done in the previous rounds. Since this probability of success (of selecting an experienced artists' painting) does not remain the same for each round, we cannot use a binomial distribution.

However, if we select the paining with replacement, that is after selecting each painting, we replace the paining and then select another one, till we select 3 paintings, then we can say that the probability of success (of selecting an experienced artists' painting) remains the same at 0.3 for each round. Now we can say that X has a binomial distribution.

Since the probability does not remain the same, we say that X has Hypergeometric distribution.

Other distributions

If X is the number of paintings selected with replacement, till we get a success ( select an experienced artists' painting), then X has a Geometric distribution. Like Binomial, the success probability needs to remain the same for each trial and there is only 1 success and the trial stops moment we have a success

If X is the number of paintings selected with replacement, till we get r successes ( select r experienced artists' painting), then X has a Negative Binomial distribution. Like Binomial, the success probability needs to remain the same for each trial and there are r successes and the trial stops moment we have r successes

For the problems with normal distribution, the mean and standard deviations are mentioned and specifically mentioned that the distribution is normal

Poisson is used when X is number of events (such as calls, arrivals etc) happen during a time period such as per minute/hour/day. When Poisson you are usually just given an average number of arrivals (say..) in a time period such as day/hour/minute etc.. Of course Poisson can be used to approximate a Binomial distribution, but then you are clear that you are asked to use a Poisson approximation to Binomial distribution.