Respuesta :
Given that we have 8 green balls and 7 orange balls, making a total of 15 balls. We can select 8 balls at random from the urn containing 15 balls in
[tex]C^n_r\text{ number of ways}[/tex]Where
[tex]C^n_r=\frac{n!}{(n-r)!r!}[/tex]n is the total outcome
r is the desired outcome
Thus, We can select 8 balls at random from the urn containing 15 balls in
[tex]C^{15}_8=\frac{15!}{(15-8)!\times8!}=\frac{15!}{7!\times8!}=\frac{15\times14\times13\times12\times11\times10\times9\times8!}{8!\times7\times6\times5\times4\times3\times2\times1}=6435\text{ ways}[/tex]We can select 3 green balls from 8 green balls (contained in the urn) at random in
[tex]C^8_3=\frac{8!}{(8-3)!\times3!}=\frac{8!}{5!\times3!}=\frac{8\times7\times6\times5!}{5!\times3\times2\times1}=56\text{ ways}[/tex]Similarly, we can select 5 orange balls from 7 orange balls (contained in the urn) at random in
[tex]C^7_5=\frac{7!}{(7-5)!\times5!}=\frac{7!}{2!\times5!}=\frac{7\times6\times5!}{5!\times2\times1}=21\text{ ways}[/tex]We can now calculate the probability of selecting 3 green balls and 5 orange balls as
[tex]P(3G,5O)=P(3G)\times P(5O)[/tex]Where P(3G) is the probability of selecting 3 green balls
P(5O) is the probability of selecting 5 orange balls
[tex]P(3G,5O)=\frac{\text{number of ways of selecting 3 green balls and 5orange balls }}{\text{number of ways of selecting 8 balls at random}}=\frac{56\text{ }\times\text{ 21}}{6435}=0.1828[/tex]
Hence, the probability of selecting 3 green balls and 5 orange balls is 0.183, in the 3 decimal places
