Question 325920: A box contains four red marbles, seven white marbles, and five blue marbles. If two marbles are drawn one at a time, find the probabilty that both marbles are white if the draws are made as folows.
(a) with replacement_______ (b) without relpacement________
Answer by Apathious(24)   (Show Source):
You can put this solution on YOUR website! Okay, this is not going to be too hard, believe me.
First, remember that probability of an event is defined to be:
P(E)= # favorable outcomes / # possible outcomes
In your problem, there are 4+7+5 = 16 marbles in the box, so in each case the # of possible outcomes is 16 to start with.
Also, when you make two draws, we have to find the probability for each draw, and multiply them together, since you want both events to happen. OK, so here we go:
With replacement:
On the first draw, there are 16 marbles in the box, and 7 of them are white, so P(White) = 7/16. Now, since you are recording your result and putting the marble back in the box, on the second draw the P(White) = 7/16 also, since the conditions are exactly the same. So in the replacement case, P(W, W) = 7/16 * 7/16 = 49/256 or 0.191
Without replacement:
On the first draw, everything is the same as the case above, so P(W) = 7/16. HOWEVER, since you are not putting the marble back in the box, on the second draw you have to assume that the first draw resulted in a favorable outcome--that is, that you drew a white marble. NOW, on the second draw, there are only 15 marbles in the bag, and only 6 of them are white, so in the case w/o replacement,
you have:
P(W, W) = 7/16 * 6/15 = 42/240 = 0.175
(In every case w/o replacement, you always assume a favorable outcome has occured on the previous draw, so for example, if you did a third draw w/o replacement, on the 3rd draw the probability of white would be 5/14, get it?)
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