An urn contains $5$red, $6$blue, and $8$green balls. If a set of $3$balls is randomly selected, what is the probability that each of the balls will be

(a) of the same color?

(b) of different colors? Repeat under the assumption that whenever a ball is selected, its color is noted and it is then replaced in the urn before the next selection. This is known as sampling with replacement .

Given in the question that there are $5$red balls, $6$blue balls,$8$green balls we have to To find the probability, if all three balls are of same color are picked and if three balls are picked with replacement from an urn containing $5$red, $6$blue, and $8$green balls.

Formula used:$P\left(A\right)=\frac{n\left(E\right)}{n\left(S\right)}$

$P\left(A\right)$is the probability of an event "A"

$n\left(E\right)$is the number of favorable outcomes.

$n\left(S\right)$is the total number of events in the sample space

Sample space is total number of possible combination of $3$balls with replacement $={19}^3$

Number of ways of selecting red balls in all three turns $5^3$

Number of ways of selecting blue balls in all three turns $6^3$

Number of ways of selecting green balls in all three turns $8^3$

Total number of ways of selecting three balls, all of the same color, in three turns is $8^3+5^3+6^3$

localid="1650281269515" role="math" $$\begin{gathered} Probabilitythatallballsareofthesamecolor=\frac{8^3+5^3+6^3}{{19}^3} \\ =0.1243 \end{gathered}$$

There are $5$red,$6$blue,$8$green balls are given. we have To find the probability, if all balls are of different color and three balls are picked with replacement from an urn containing $5$red, $6$blue, and $8$green balls.

Formula used $P\left(A\right)=\frac{n\left(E\right)}{n\left(S\right)}$

Number of ways to select balls of different color in $3$turn is$8\times 6\times 5\times 3!$

localid="1650281229902" $$\begin{gathered} Probabilitythatallballsareofdifferentcolor=\frac{8\times 6\times 5\times 3!}{1963} \\ =0.7335 \end{gathered}$$