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A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
Answer & Solution
Correct Answer:
Option
A
Total number of balls
= (2 + 3 + 2)
= 7
Let S be the sample space
Then, n(S) = Number of ways of drawing 2 balls out of 7
$$\eqalign{ & {\text{n}}\left( {\text{S}} \right) = {}^7{C_2} \cr & {\text{n}}\left( {\text{S}} \right) = \frac{{\left( {7 \times 6} \right)}}{{\left( {2 \times 1} \right)}} \cr & {\text{n}}\left( {\text{S}} \right) = 21 \cr} $$
Let E = Event of 2 balls, none of which is blue
∴ n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls
$$\eqalign{ & {\text{n}}\left( {\text{E}} \right)\, = {}^5{C_2} \cr & {\text{n}}\left( {\text{E}} \right) = \frac{{\left( {5 \times 4} \right)}}{{\left( {2 \times 1} \right)}} \cr & {\text{n}}\left( {\text{E}} \right) = 10 \cr & \therefore {\text{P}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} = \frac{{10}}{{21}} \cr} $$
= (2 + 3 + 2)
= 7
Let S be the sample space
Then, n(S) = Number of ways of drawing 2 balls out of 7
$$\eqalign{ & {\text{n}}\left( {\text{S}} \right) = {}^7{C_2} \cr & {\text{n}}\left( {\text{S}} \right) = \frac{{\left( {7 \times 6} \right)}}{{\left( {2 \times 1} \right)}} \cr & {\text{n}}\left( {\text{S}} \right) = 21 \cr} $$
Let E = Event of 2 balls, none of which is blue
∴ n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls
$$\eqalign{ & {\text{n}}\left( {\text{E}} \right)\, = {}^5{C_2} \cr & {\text{n}}\left( {\text{E}} \right) = \frac{{\left( {5 \times 4} \right)}}{{\left( {2 \times 1} \right)}} \cr & {\text{n}}\left( {\text{E}} \right) = 10 \cr & \therefore {\text{P}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} = \frac{{10}}{{21}} \cr} $$
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Loginprobability of first ball being blue=2/7
its complementary =1-(2/7)
same for second ball is blue p(b|a)=2/6
its complementary =1-(2/6)
now {1-(2/7)}*{1-(2/6)}
=10/21