A team of 8 couples attend a lucky draw in which 4 persons are picked for a prize. Then the probability that there is at least one couple is

A. 11/39 B. 12/39 C.14/39 D. 15/39

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Solution

Option D is correct

there are 8 couples...i.e. 16 persons...
so total outcomes for choosing 4 persons is 16C4 = 1820
p(atleast 1 couple) = p(1 couple or 2couples)
for 1 couple -> we have 8 choices to choose a couple,
for remaining 2 non couple persons, we can select any 1 person from 14 people in 14 ways
now the 4th person can't be spouse of 3rd person, so we have only 12 persons to choose from
so total choices for 3rd and 4th persons are 1412/2 (divided by 2 as we are counting a pair twice) and total choices including 1st and 2nd persons are 814*12/2
which is 672
Now for 2 couples we have 8c2 = 28 choices
so total favuorable outcomes = 672 + 28 = 700
700/1820 = 5/13, or 15/39

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$(\frac{7}{- 26} + \frac{16}{39}) = ?$

(a) $\frac{11}{78}$

(b) $\frac{- 11}{78}$

(d) $\frac{- 11}{39}$

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