Q. 2.17 - A First Course in Probability

Question

Five balls are randomly chosen, without replacement, from an urn that contains $5$red,$6$ white, and $7$blue balls. Find the probability that at least one ball of each color is chosen.

Step-by-Step Solution

Answer

$P\left(A\right)=\frac{287}{408}\approx 0.70343$$\frac{287}{408}\approx 0.70343$

It is simpler to count the probability of the complement, which can be shown as a union of three mutually exclusive events.

Step 1 Given information.

Five balls are randomly chosen, without replacement, from an urn that contains$5$ red, $6$white, and $7$blue balls.

Step 2 Explanation.

The described experiment is equivalent to:

The experiment: From an urn that contains $5$red balls, $6$white balls, and $7$blue ones, draw $5$balls at random

Find the Probability that all three colors appear in $5$chosen balls.

The outcome space of the experiment $S$contains all of the combinations (subsets of fixed size) of $5$different balls.

If all events are considered equally likely, the probability of an event $A\subseteq S$ is:

$P\left(A\right)=\frac{\left|A\right|}{\left|S\right|}$

where $\left|X\right|$denotes the number of elements in$X$.

It is shown that the number of combinations$5$ of$18$ different objects is$\left(\begin{array}{c}18\\ 5\end{array}\right)=\left|S\right|$.

Step 3 Explanation.

Say that $A$is an event in that balls of all three colors are chosen.

Since direct enumeration would have to have many cases, we count elements$A^c$.

$P\left(A\right)=1-P\left(A^c\right)$

$A^c$is a union of three disjoint events:

That only blue and red balls are chosen - $\left(\begin{array}{c}12\\ 5\end{array}\right)$possible choices of$5$ out of those$12$ balls.

That only blue and white balls are chosen - $\left(\begin{array}{c}13\\ 5\end{array}\right)$possible choices.

That only red and white balls are chosen - $\left(\begin{array}{c}11\\ 5\end{array}\right)$possible choices.

Total of $\left(\begin{array}{c}11\\ 5\end{array}\right)+\left(\begin{array}{c}12\\ 5\end{array}\right)+\left(\begin{array}{c}13\\ 5\end{array}\right)$elements in$A^c$.

So$P\left(A^c\right)=\frac{\left(\begin{array}{c}11\\ 5\end{array}\right)+\left(\begin{array}{c}12\\ 5\end{array}\right)+\left(\begin{array}{c}13\\ 5\end{array}\right)}{\left(\begin{array}{c}18\\ 5\end{array}\right)}=\frac{121}{408}$

$P\left(A\right)=1-P\left(A^c\right)=1-\frac{121}{408}=\frac{287}{408}\approx 0.70343$

The end result is:

$P\left(A\right)=\frac{287}{408}\approx 70.34\%$.