1. There are 6 white, 8 green and 12 red balls in a bag. Two balls are drawn at random with replacement. what is the probability that
(i) both are red?
(ii) both are of the same colour
(iii) they are different in colour?
2.The probability that a man will be alive in 25 years is 0.75, and the probability that his wife will be alive in 25 years is 0.88. Find the probability that
(i) both will be alive in 25 years;
(ii) only his wife will be alive in 25years;
(iii) at least one will be alive in 25 years.
Probabilities thanks
2015-04-28 5:49 pm
回答 (2)
2015-04-28 9:41 pm
✔ 最佳答案
1. i) P(R^R)=12/26*12/26
ii) P(same color)=(6C2+8C2+12C2)/26C2
iii) P(different) =1 minus above prob
2.
i) P(both live)=.75*.88
ii) P(wife live)=(1-.75)*.88
iii) P(at least one live)=1-P(both die)=1-(1-.75)*(1-.88)
Correct me if it's wrong. Kinda sleepy now. G'night.
2015-04-29 10:42:51 補充：
Exactly. You are right pingshek.
參考: myself~
2015-04-29 8:32 am
1.(ii)
Pr(same colour)
= Pr(both are white) + Pr(both are green) + Pr(both are red)
= (6/26)² + (8/26)² + (36/169)
= (9/169) + (16/169) + (36/169)
= 61/169
Star's answer is for two balls drawn WITHOUT replacement.
Pr(same colour)
= Pr(both are white) + Pr(both are green) + Pr(both are red)
= (6/26)² + (8/26)² + (36/169)
= (9/169) + (16/169) + (36/169)
= 61/169
Star's answer is for two balls drawn WITHOUT replacement.
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