Plzz
a package contains 4 red, 2 green, 8 purple, and 6 blue jelly beans. what is the probability of choosing a purple jelly bean, eating it, and then choosing a blue jelly bean?

There's one piece of information missing from the question, so
in order to completely calculate the answer, we must assume the
missing number.

We have no information that helps us calculate the probability
that you will eat the first jelly bean.  So we'll have to assume that
you will definitely eat it, no matter what color it is.  The probability is 1 .

-- You start out with  (2 + 4 + 8 + 6) = 20 jelly beans in the package.
    8 of them are purple.

-- If you close your eyes and pull one out of the package, then the
probability that it will be purple is
                                                  8 / 20 .

-- You definitely eat it.  The probability of this event is 1 .

-- Now there are  19  jelly beans left in the package.
   6 of them are blue.

-- The probability of picking a blue one out of the package without peaking is

                             6 / 19 .

-- The probability of all three of these events happening is

                       (8/20) · (1) · (6/19)

                   =  (8 · 1 · 6) / (20 ·19)

                   =      48 / 380 

                   =  12/95  =  0.1263...  =  12.63...%  (rounded)   

Ithink a hope that

7/12 is the ¡!