Question

Two balls are drawn in succession without replacement from an urn containing 5 red balls and 6 blue balls. Let Z be the random variable representing the number of blue balls. Construct the probability distribution and histogram of the random variable Z.

Answer 1

ANSWER and EXPLANATION

Let R represent the number of red balls.

Let B represent the number of blue balls.

There are four possible outcomes when the balls are picked:

`\lbrace RR,RB,BR,BB\rbrace`

We have that Z is the random variable that represents the number of blue balls.

This implies that the possible values of Z are:

To construct the probability distribution, we have to find the probabilities of each of the outcomes:

`\begin{gathered} P(RR)=(5)/(11)*(4)/(10)=(2)/(11) \\ P(RB)=(5)/(11)*(6)/(10)=(3)/(11) \\ P(BR)=(5)/(11)*(6)/(10)=(3)/(11) \\ P(BB)=(6)/(11)*(5)/(10)=(3)/(11) \end{gathered}`

Hence, the probabilities for the possible outcomes of the random variable are:

`\begin{gathered} P(Z=0)=(2)/(11) \\ P(Z=1)=(3)/(11)+(3)/(11)=(6)/(11) \\ P(Z=2)=(3)/(11) \end{gathered}`

Therefore, the probability distribution is:

Now, let us plot the histogram:

That is the answer.