Question
A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If two marbles are drawn at random, what is the probability that at least one is green?
Answer: Option A
Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
Probability that at least one green marble can be picked in the random draw of two marbles = Probability that one is green + Probability that both are green
$$\eqalign{
& = \frac{{{}^6{C_1}\, \times \,{}^9{C_1}}}{{{}^{15}{C_2}}} + \frac{{{}^6{C_2}}}{{{}^{15}{C_2}}} \cr
& = \frac{{6 \times 9 \times 2}}{{15 \times 14}} + \frac{{6 \times 5}}{{15 \times 14}} \cr
& = \frac{{36}}{{70}} + \frac{1}{7} \cr
& = \frac{{46}}{{70}} \cr
& = \frac{{23}}{{35}} \cr} $$
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Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
Probability that at least one green marble can be picked in the random draw of two marbles = Probability that one is green + Probability that both are green
$$\eqalign{
& = \frac{{{}^6{C_1}\, \times \,{}^9{C_1}}}{{{}^{15}{C_2}}} + \frac{{{}^6{C_2}}}{{{}^{15}{C_2}}} \cr
& = \frac{{6 \times 9 \times 2}}{{15 \times 14}} + \frac{{6 \times 5}}{{15 \times 14}} \cr
& = \frac{{36}}{{70}} + \frac{1}{7} \cr
& = \frac{{46}}{{70}} \cr
& = \frac{{23}}{{35}} \cr} $$
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