Question 1162401: We are gives two urns follows:
Un A contains 6 red marbles, 2 white marbles and 3 blue marbles.
Urn B contains 4 red marbles, 5 white marbles and 1 blue marble.
A fair die is tossed; if1 or 5 appears, a marble is chosen from urn B,
otherwise a marbles is chosen from urn A. find the probability that:
A red marble is chosen.
A white marble is chosen.
A blue marble is chosen.
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
Note that the marble that is chosen is either red, white, or blue. So the sum of the probabilities should be 1. Use that to check the reasonableness of your calculations.
P(A) = 2/6 = 1/3
P(B) = 4/6 = 2/3
P(A,red) = (1/3)(6/11) = 6/33
P(A,white) = (1/3)(2/11) = 2/33
P(A,blue) = (1/3)(3/11) = 3/33
P(B,red) = (2/3)(4/10) = 8/30
P(B,white) = (2/3)(5/10) = 10/30
P(B,blue) = (2/3)(1/10) = 2/30
ANSWERS:
P(red) = 6/33+8/30 = 2/11+4/15 = 30/165+44/165 = 74/165
P(white) = 2/33+10/30 = 2/33+1/3 = 2/33+11/33 = 13/33 = 65/165
P(blue) = 3/33+2/30 = 1/11+1/15 = 15/165+11/165 = 26/165
CHECK: 74/165+65/165+26/165 = 165/165 = 1
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