What makes this question messy is the fact that the bags contain different number of balls, so we have to take all cases
BagA:
WW - 2/6 * 1/5 = 2/30 = 1/15
WB - 2/6 * 1/5 * 2 = 4/30 = 2/15 , times 2 since it could be BW
WR - 2/6 * 3/5 * 2 = 12/30 =6/15
BR - 1/6 * 3/5 * 2 = 6/30 = 3/15
RR - 3/6 *2/5 = 6/30 = 3/15
BB - not possible, there is only 1 Black
notice the sum is 15/15 , all accounted for
Bag B:
WW - 3/9 * 2/8 = 6/72 = 1/12 = 3/36
WB - 3/9 * 2/8 * 2 = 12/72 = 1/6 = 6/36
WR - 3/9 * 4/8 * 2 = 24/72 = 1/3 = 12/36
BR - 2/9 * 4/8 * 2 = 16/72 = 2/9 = 8/36 = 40/180
RR - 4/9 * 3/8 = 1/6 = 6/36
BB - 2/9 * 1/8 = 1/36
Again, note the sum is 36/36
Bag C
WW = 4/9 * 3/8 = 6/36
WB - 4/9 * 3/8 * 2 = 12/36
WR - 4/9 * 2/8 * 2 = 8/36
BR - 3/9 * 2/8 * 2 = 6/36
RR - 3/9 * 2/8 = 3/36
BB - 2/9 * 1/8 * 2 = 1/36
It would be nice if they all had the same common denominator
which would be 180
So if we change all the fractions above out of 180, we would have 540 in total,
the one we need is BagB,RB = 8/36 = 40/180
so the prob of your event is 40/540 = 2/27
Bag A contains 2 white, 1 black and 3 red balls, Bag B contains 3 white, 2 black and 4 red balls and Bag C contains 4 white, 3 black and 2 red balls. One Bag is chosen at random and 2 balls are drawn at random from that Bag. Of the randomly drawn balls happen to be red and black, what is the probability that both balls come from Bag B?
1 answer
1 answer