Question from http://www.sgforums.com/forums/2297/topics/314258
1) Two boxes are labelled A and B. box A contain 3 blue discs and 2 white discs and box B contain 2 blue discs and 3 white discs. A random sample of 2 discs is selected without replacement from each box. Find the probability that at most one of the discs drawn is white.
2) Two friends Jane and Mary often go their shopping on sat afternoon. If Mary goes shopping, the probability that Jane also goes is 0.96. If Jane goes, the probability that Mary goes is 0.8. The probability that neither goes is 0.07. Find the probability that both go shopping.
Answer:
1)
P(none white) = 3/5 * 2/4 * 2/5 * 1/4 = 12/400 = 3/100
P(1 white)
= P(1 white from box A and no white from box B) + P( no white from box A and 1 white from box B)
= ( 2/5 *3/4 *2! ) * (2/5*1/4) + (3/5 * 2/4) * (3/5 * 2/4 * 2!)
= 3/50 + 9/50 = 12/50
Summing up gives 27/100
2)
P(at least one of them go) = P(J goes) + P(M goes) - P(both J and M goes) {double counted}
1 - P(none go) = P(both J and M goes)/0.96 + P(both J and M goes)/0.8 - P(both J and M goes)
1-0.07 = P(both J and M goes)[1/0.96+1/0.8-1]
P(both J and M goes) = 0.72
(I'm not very sure of Question 2, will review it again later... If you have any ideas, please email me. Thanks!)
