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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?
- A14/33
- B9/22
- C1/6
- D1/3
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Discussion 1
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CCommunity
⇒ 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
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