your first answer is the probability of drawing RRB in that order. But, if all you want is two reds and a blue, you need to multiply that by 3!/1!2!=3 because there are that many indistinguishable ways to reorder the balls. In other words, you have
RRB, RBR, BRR
They are all equally likely to occur.
Your second answer is correct, since both balls are the same.
A jar has 25 red, 15 blue and 10 green marbles. Drawing 3 marbles all at once what is the probability of drawing 2 red and 1 blue? 2 red? Not sure if this is right.
For the fist problem I multiplied (25/50)*(24/49)*(15/48)=.0765
and for the next problem I multiply (25/50)*(24/49)*(25/48)=.128
1 answer