232nd
P(required) = 1- P( no pairs + 4 pairs)
no pairs = (2C1)8 = 28
4 pairs = 8C4
so P = 1- (28+8C4)16C8
I am confused in the approach to two similar questions. So I am posting both-
1) A team of 8 couples (husband and wife) attend a lucky draw in which 4 persons are picked up for a prize. The probability that there is atleast one couple is?
2) 8 pairs of shoes are in a closet. 8 shoes are selected at random. The probability that there will be at least one pair and atmost 3 pairs among selected shoes is?
Attempt:
1) Required probability = 1 - probability of not selecting any couple for prize
= 1 - (16 x 14 x 12 x 10)/(16P4)
= 5/13
2) Required probability = 1 - probability of selecting no pair - probability of selecting 4 pairs
= 1 - (16 x 14 x 12 x 10 .....x 2)/(16P8) - (16 x 1 x 14 x 1.....10 x 1)/(16P8)
= 344/351
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UP 0 DOWN 0 0 5
5 Answers
491 is ok
2) P(at least 1 pair at most 3 pairs)=1-P(no pair)-P(4 pairs)-P(5pairs) - P(6 pairs) - P(7 pairs) -P(8 pairs) [1]
11 is not ok [4]
The answer is 15/39
2) only 8 shoes are selected so maximum are 4 pairs. No posiiblity of 5,6,7....[1]
for this the answer is 1 - 326/16C8
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