A jar holds 1 red, 1 green, 1 blue, and 1 yellow gumball.The gumballs are
selected from the jar in random order. If the first gumball is red, what is the
probability that the yellow gumball is selected before the blue gumball?
Method 1 (the easier way):
It is just as likely that the yellow gumball will be drawn before the blue one
as it is that the blue one will be drawin before the yellow one. So the
desired probability is .
That is, in one half of the orders of drawing, the yellow gumball will come
before the blue and in the other half of the orders of drawing, the blue one
will come before the yellow one, so the probability is .
Method 2:
Since you are given that the first one is red, then the sample space is
this set of orders of gumballs drawn:
{ RGBY, RGYB, RBGY, RBYG, RYGB, RYBG}, that's 6 possible ways.
The "success" space is the set where the the yellow gumball is selected before
the blue gumball:
{ RGYB, RYGB, RYBG}, that's 3 successful ways.
So the probability is 3 out of 6 or or .
Edwin