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?
A. $$\frac{{10}}{{21}}$$
B. $$\frac{{11}}{{21}}$$
C. $$\frac{{2}}{{7}}$$
D. $$\frac{{5}}{{7}}$$
Answer: Option A
Solution(By Examveda Team)
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
$$\eqalign{ & {\text{n}}\left( {\text{E}} \right)\, = {}^5{C_2} \cr & {\text{n}}\left( {\text{E}} \right) = \frac{{\left( {5 \times 4} \right)}}{{\left( {2 \times 1} \right)}} \cr & {\text{n}}\left( {\text{E}} \right) = 10 \cr & \therefore {\text{P}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} = \frac{{10}}{{21}} \cr} $$
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Comments ( 6 )
Related Questions on Probability
A. $$\frac{{1}}{{2}}$$
B. $$\frac{{2}}{{5}}$$
C. $$\frac{{8}}{{15}}$$
D. $$\frac{{9}}{{20}}$$
A. $$\frac{{10}}{{21}}$$
B. $$\frac{{11}}{{21}}$$
C. $$\frac{{2}}{{7}}$$
D. $$\frac{{5}}{{7}}$$
A. $$\frac{{1}}{{3}}$$
B. $$\frac{{3}}{{4}}$$
C. $$\frac{{7}}{{19}}$$
D. $$\frac{{8}}{{21}}$$
E. $$\frac{{9}}{{21}}$$
What is the probability of getting a sum 9 from two throws of a dice?
A. $$\frac{{1}}{{6}}$$
B. $$\frac{{1}}{{8}}$$
C. $$\frac{{1}}{{9}}$$
D. $$\frac{{1}}{{12}}$$
we will apply the complementary rule p(a)=1-p(a')
probability 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
1 - 2C1×5C1+2C2/7C2
2C1×5C1+2C2/7C2
Why you have not included another case, where 1 ball can be blue and and another ball can be picked from other 5 balls.
Thanks 😊👍 for the answer but can we solve this without using permutations and combinations??if yes then please tell us
where is equal sign before 10 in the 2nd last line?