Question

Four red, $8$blue, and $5$green balls are randomly arranged in a line.

$\left(a\right)$What is the probability that the first $5$balls are blue?

$\left(b\right)$What is the probability that none of the first $5$balls is blue?

$\left(c\right)$What is the probability that the final $3$balls are of different colors?

$\left(d\right)$What is the probability that all the red balls are together?

Short answer

$a)$width="204" height="21" role="math">P(thefirst5ballsareblue)$=\frac{2}{221}\approx 0.009$.

$b)$role="math" localid="1649489349358" $P(noneofthefirst5ballsareblue)$$=\frac{9}{442}\approx 0.02$.

$c)$$P(thefinalthreeballsareofdifferentcolors)$$=\frac{4}{17}\approx 0.24$.

$d)$$P(alltheredballsaretogether)$$=\frac{1}{34}\approx 0.03$.

Step-by-step solution

Given Information.

Four red, $8$blue, and $5$green balls are randomly arranged in a line.

Explanation.

The described experiment is equivalent to:

The experiment: randomly permute (set in a line) a group of $4$red, $8$blue, and $5$green balls.

The outcome space of the experiment $S$contains all permutations of the$17$different objects.

If all events $S$are considered equally likely, the probability of the 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$X$From the basic principle of counting $\left|S\right|=17!$

part (a) Explanation.

$a)$$P$(the first $5$balls are blue)

Count the number of elements in this event (which are members of$S$).

There are $8$choices for the first blue ball, $7$choices for the second ball (of the remaining blue balls), etc.

There are $8·7·6·5·4$different choices for the first $5$balls, and the remaining $12$balls can be permuted in any of the$12!$ways

$P\left(\text{the first}5\text{balls are blue}\right)$$=\frac{8·7·6·5·4·12!}{17!}$$=\frac{2}{221}\approx 0.009$.

part (b) Explanation.

$b)$$P$(none of the first $5$balls are blue)

Count the number of elements $S$that are in this event.

There are $9$balls that aren't blue, $8$choices for the second ball after the first nonblue ball is chosen, etc.

There are $9·8·7·6·5$different choices for the first $5$balls, and the remaining $12$balls can be permuted in any of therole="math" localid="1649488044793" $12!$ways

$P\left(\text{none of the first}5\text{balls are blue}\right)$$=\frac{9·8·7·6·5·12!}{17!}$$=\frac{9}{442}\approx 0.02$

part (c) Explanation.

$c)$width="13" height="19" role="math">P( the final three balls are of different colors)

Count the number of permutations in this event. It does not matter from which end the counting starts as long as the choice of some ball does not affect the number of choices for the balls in the remaining places.

Choose the red ball that will be in the final three among $4$the red balls$-4$results, then choose the blue ball in $8$ways, and the green ball in$5$ways. For every such choice, different permutations of these three balls create different elements$S$. There are $4·8·5·3!$different possibilities for the three final balls.

For every permutation of the final balls, the remaining $14$balls can be permuted in any of the$14!$ways.

Now use the basic principle of counting.

$P\left(\text{the final three balls are of different colors}\right)$$=\frac{4·8·5·3!·14!}{17!}$$=\frac{4}{17}\approx 0.24$

part (d) Explanation.

$d)$$P$( all the red balls are together)

Count the number of permutations in this event.

Consider the red balls as one element. Then there are $14$elements, and there are $14!$permutations of those. However$S$, the order of the red balls is also included, so there are in fact localid="1649489123503" $5!$times more elements in this event.

$P\left(\text{all the red balls are together}\right)$$=\frac{5!·14!}{17!}$localid="1649489786537" $=\frac{1}{34}\approx 0.03$