Question
A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:
Answer: Option C
Let S be the sample space
Then, n(S) = number of ways of drawing 3 balls out of 15
$$\eqalign{
& = {}^{15}{C_3} \cr
& = \frac{{ {15 \times 14 \times 13} }}{{ {3 \times 2 \times 1} }} \cr
& = 455 \cr} $$
Let E = event of getting all the 3 red balls
$$\eqalign{
& \therefore n\left( E \right) = {}^5{C_3} = {}^5{C_2} \cr
& = \frac{{ {5 \times 4} }}{{ {2 \times 1} }} = 10 \cr
& \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr
& = \frac{{10}}{{455}} \cr
& = \frac{2}{{91}} \cr} $$
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Let S be the sample space
Then, n(S) = number of ways of drawing 3 balls out of 15
$$\eqalign{
& = {}^{15}{C_3} \cr
& = \frac{{ {15 \times 14 \times 13} }}{{ {3 \times 2 \times 1} }} \cr
& = 455 \cr} $$
Let E = event of getting all the 3 red balls
$$\eqalign{
& \therefore n\left( E \right) = {}^5{C_3} = {}^5{C_2} \cr
& = \frac{{ {5 \times 4} }}{{ {2 \times 1} }} = 10 \cr
& \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr
& = \frac{{10}}{{455}} \cr
& = \frac{2}{{91}} \cr} $$
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