Answer-
a. Probability that  three of the candies are white = 0.29
b. Probability that three are white, 2 are tan, 1 is pink, 1 is yellow, and 2 are green = 0.006
Solution-
There are 19 white candies, out off which we have to choose 3.
The number of ways we can do the same process =
[tex]\binom{19}{3} = \frac{19!}{3!16!} = 969[/tex]
As we have to draw total of 9 candies, after 3 white candies we left with 9-3 = 6, candies. And those 6 candies have to be selected from 52-19 = 33 candies, (as we are drawing candies other than white, so it is subtracted)
And this process can be done in,
[tex]\binom{33}{6} = \frac{33!}{6!27!} =1107568[/tex]
So total number of selection = (969)×(1107568) = 1073233392
Drawing 9 candies out of 52 candies,
[tex]\binom{52}{9} = \frac{52!}{9!43!} = 3679075400[/tex]
∴P(3 white candies) = [tex]\frac{1073233392}{3679075400} =0.29[/tex]
Total number of ways of selecting 3 whites, 2 are tans, 1 is pink, 1 is yellow, and 2 are greens is,
[tex]\binom{19}{3} \binom{10}{2} \binom{7}{1} \binom{5}{1} \binom{6}{2}[/tex]
[tex]=(\frac{19!}{3!16!}) (\frac{10!}{2!8!}) (\frac{7!}{1!6}) (\frac{5!}{1!4!}) (\frac{6!}{2!4!})[/tex]
[tex]=(969)(45)(7)(5)(15)=22892625[/tex]
Total number of selection = 3 whites + 2 are tans + 1 is pink + 1 is yellow + 2 greens = 9 candies out of 52 candies is,
[tex]\binom{52}{9}=\frac{52!}{9!43!} =3679075400[/tex]
∴ P( 3 whites, 2 are tans, 1 is pink, 1 is yellow, 2 greens) =
[tex]\frac{22892625}{3679075400} = 0.006[/tex]
