Question

Suppose we have an urn containing 9 marbles. Two are red, three are green, and four are blue. We randomly select 5 marbles from the urn, with replacement. What is the probability of selecting 3 green marbles, 1 red marble, and 1 blue marbles?

Answer 1

Answer:

probability of selecting 3 green marbles, 1 red marble, and 1 blue marbles

$P(E) =\frac{8}{126} = 0.0634$

Step-by-step explanation:

Given data urn containing '9' marbles

given  red marbles = 2

         green marbles =3

          blue marbles = 4

Five marbles can be selected at a time from '9' marbles in $9_{C_{5} }$   ways

by using  formula $n_{C_{r} }=\frac{n!}{(n-r)!r!}$

                          $9_{C_{5} }=\frac{9!}{(9-5)!5!} = \frac{9X8X7X6X5!}{4!5!}$

After simplification , we get $9_{C_{5} } = 126$

         total number of ways n(S) = 126

The probability of selecting '3' green marbles ,one red marble and one blue marble with replacement.

let ' E' be the event of selecting '3' green marbles ,one red marble and one blue marble with replacement.

n(E) = $3_{C_{3} } X2_{C_{1} }X 4_{C_{1} }$  ways

The required probability $P(E) = \frac{n(E)}{n(S)}$

$p(E) = \frac{3_{C_{3} } X2_{C_{1} }X 4_{C_{1} }}{9_{C_{5} } }$

on simplification , we get

$P(E) = \frac{1X2X4}{126} =\frac{8}{126}$

$P(E) =\frac{8}{126} = 0.0634$