A room contains 3 red, 5 green and 4 blue chairs. Two chairs are picked and are put in the lawn. What is the probability that none of the chairs picked is blue?
Correct Answer: Option A
⇒ Total number of chairs = (3 + 5 + 4) = 12.
Let S be the sample space.
Then, n(s)= Number of ways of picking 2 chairs out of 12
\({ = ^{12}}{C_2} = \frac{{12 \times 11}}{{2 \times 1}} = {\rm{}}66\)
Let n(E) = number of events of selecting 2 chairs for selecting no blue chairs.
\(= {{\rm{}}^8}{{\rm{C}}_2} = \frac{{8 \times 7}}{{2 \times 1}} = 28\)
Therefore required probability = 28/66 = 14/33
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