Answer:[tex] \frac{5}{91},\frac{38}{182} [/tex]
Step-by-step explanation:
(a)Probability that these three marbles are all red
choosing 3 red out of 6 is given by [tex]^6C_3[/tex]
and choosing 3 marble from 14 marbles=[tex]^{14}C_3[/tex]
[tex]Probability=\frac{^6C_3}{^{14}C_3}=\frac{5}{91}[/tex]
(b)All of same color
It is possible when they are of either red or green
For red it is [tex]\frac{^6C_3}{^{14}C_3}[/tex]
For Green it is
[tex]\frac{^8C_3}{^{14}C_3}[/tex]
total=[tex]\frac{^6C_3}{^{14}C_3}+\frac{^6C_3}{^{14}C_3}[/tex]
=[tex] \frac{38}{182} [/tex]
