Question
In a party there are 5 couples. Out of them 5 people are chosen at random. Find the probability that there are at the least two couples?
Answer: Option A
Number of ways of (selecting at least two couples among five people selected) = $$\left( {{}^5{C_2} \times {}^6{C_1}} \right)$$
As remaining person can be any one among three couples left.
Required probability
$$\eqalign{
& = \frac{{{}^5{C_2} \times {}^6{C_1}}}{{{}^{10}{C_5}}} \cr
& = \frac{{\left( {10 \times 6} \right)}}{{252}} \cr
& = \frac{5}{{21}} \cr} $$
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Number of ways of (selecting at least two couples among five people selected) = $$\left( {{}^5{C_2} \times {}^6{C_1}} \right)$$
As remaining person can be any one among three couples left.
Required probability
$$\eqalign{
& = \frac{{{}^5{C_2} \times {}^6{C_1}}}{{{}^{10}{C_5}}} \cr
& = \frac{{\left( {10 \times 6} \right)}}{{252}} \cr
& = \frac{5}{{21}} \cr} $$
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