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## Homework Statement

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?

## The Attempt at a Solution

1) Required probability = 1 - probability of not selecting any couple for prize

= 1 - (16 x 14 x 12 x 10)/(

^{16}P

_{4})

= 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)/(

^{16}P

_{8}) - (16 x 1 x 14 x 1...10 x 1)/(

^{16}P

_{8})

= 344/351

Is the method right?

1)15/39 2)1 - 326/

^{16}C_{8}
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