Aptitude  Probability  Discussion
Discussion Forum : Probability  General Questions (Q.No. 2)
2.
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: Option
Explanation:
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  
= ^{7}C_{2} `  


= 21. 
Let E = Event of drawing 2 balls, none of which is blue.
n(E)  = Number of ways of drawing 2 balls out of (2 + 3) balls.  
= ^{5}C_{2}  


= 10. 
P(E) =  n(E)  =  10  . 
n(S)  21 
Discussion:
106 comments Page 1 of 11.
Somesh Saurabh said:
7 years ago
See there are two ways to solve this question:
METHOD 1:
Probability of getting no blue balls in 2 draws = probability of getting 2 other balls in 2 draws out of any other 5 balls.
i.e., select two balls from the remaining 5 balls in 5C2 ways = n(E),
Now the sample space is n(S) = selection of 2 balls out 7b balls = 7C2,
So, required probability= n(E)/n(S),
=> 5C2/7C2 = [5!/(2!*3!)]/[7!/(2!*5!)]
=> 10/21 is the answer.
METHOD 2:
Probability of not getting any blue balls in two draws = 1  (probability of getting 1 blue ball and 1 other ball + probability of getting 2 blue balls)
So,
Selection of 1 blue ball and one other ball = 2C1 * 5C1 = 10 ways,
Selection of 2 blue balls = 2C2 = 1,
So, probability = 1  [{10/7C2} + {1/7C2}]
=> 10/21 answer.
METHOD 1:
Probability of getting no blue balls in 2 draws = probability of getting 2 other balls in 2 draws out of any other 5 balls.
i.e., select two balls from the remaining 5 balls in 5C2 ways = n(E),
Now the sample space is n(S) = selection of 2 balls out 7b balls = 7C2,
So, required probability= n(E)/n(S),
=> 5C2/7C2 = [5!/(2!*3!)]/[7!/(2!*5!)]
=> 10/21 is the answer.
METHOD 2:
Probability of not getting any blue balls in two draws = 1  (probability of getting 1 blue ball and 1 other ball + probability of getting 2 blue balls)
So,
Selection of 1 blue ball and one other ball = 2C1 * 5C1 = 10 ways,
Selection of 2 blue balls = 2C2 = 1,
So, probability = 1  [{10/7C2} + {1/7C2}]
=> 10/21 answer.
Magnus The Great said:
5 years ago
@ALL.
According to me,
2 red
3 green
2 blue
Total Balls =7 Balls.
Note: (Always solve as *without replacement* unless otherwise stated)
Pr(1st Red & 2nd red...RR) = 2/7 * 1/6 = 1/21,
Pr(1st Green & 2nd Green..RG)=3/7*2/6= 1/7,
Pr( 1st Blue & 2nd blue..BB) = 2/7 * 1/6 = 1/21,
Pr(1st Red & 2nd Green...RG) = 2/7 * 3/6 = 1/7,
Pr(1st Red & 2nd blue...RB) = 2/7 * 2/6 = 2/21,
Pr(1st Green & 2nd blue...GB) = 3/7 * 2/6 = 1/7,
Pr(1st green & 2nd red ...GR) = 3/7 * 2/6 = 1/7,
Pr(1st blue & 2nd red.... BR) = 2/7 * 1/6 = 1/21,
Pr(1st blue & 2nd green...BG) = 2/7 * 3/6 = 1/7.
THEN
Pr( None is Blue) = RR or GG or RG or GR.
1/21 + 1/7 + 1/7 + 1/7.
= 10/21.
According to me,
2 red
3 green
2 blue
Total Balls =7 Balls.
Note: (Always solve as *without replacement* unless otherwise stated)
Pr(1st Red & 2nd red...RR) = 2/7 * 1/6 = 1/21,
Pr(1st Green & 2nd Green..RG)=3/7*2/6= 1/7,
Pr( 1st Blue & 2nd blue..BB) = 2/7 * 1/6 = 1/21,
Pr(1st Red & 2nd Green...RG) = 2/7 * 3/6 = 1/7,
Pr(1st Red & 2nd blue...RB) = 2/7 * 2/6 = 2/21,
Pr(1st Green & 2nd blue...GB) = 3/7 * 2/6 = 1/7,
Pr(1st green & 2nd red ...GR) = 3/7 * 2/6 = 1/7,
Pr(1st blue & 2nd red.... BR) = 2/7 * 1/6 = 1/21,
Pr(1st blue & 2nd green...BG) = 2/7 * 3/6 = 1/7.
THEN
Pr( None is Blue) = RR or GG or RG or GR.
1/21 + 1/7 + 1/7 + 1/7.
= 10/21.
Sandeep said:
10 years ago
If you are not aware of combinations like 7C2 then give names to the balls like r1, r2, g1, g2, g3, b1 and b2.
So if you take two balls out of the bag then if we consider r1 comes then total combination would be r1r2, r1g1, r1g2.. and so on. That means, for r1 there are 6 combinations.
So we can conclude following table.
Ball combination
r1 6
r2 5
g1 4
g2 3
g3 2
b1 1
b2 0
So total number of combination = 6+5+4+3+2+1 = 21.
And combination using blue ball = 2+2+2+2+2+1 = 11.
So combination without using blue ball = 2111 = 10.
So probability of the same = 10/21 :).
So if you take two balls out of the bag then if we consider r1 comes then total combination would be r1r2, r1g1, r1g2.. and so on. That means, for r1 there are 6 combinations.
So we can conclude following table.
Ball combination
r1 6
r2 5
g1 4
g2 3
g3 2
b1 1
b2 0
So total number of combination = 6+5+4+3+2+1 = 21.
And combination using blue ball = 2+2+2+2+2+1 = 11.
So combination without using blue ball = 2111 = 10.
So probability of the same = 10/21 :).
Precious said:
6 years ago
When we say 6C subscript 2, it means
6 * 5 * 4 * 3 * 2 * 1 divided by 2 * 1.
So, the correct answer to me is 5/7.
Since, 2 red, 3 green, and 2 blue, are present,
Therefore,
S=7 (numbers of the balls)
Therefore,
Let E be the event space that blue ball is been chosen at random.
n(E)=2
n(s)=7 (total numbers of balls in the bag.
Pr(E)=n(E)/n(s)
=2/7.
To find the probability that none of the blue ball is drawn, 1 must be subtracted from the probability that the blue ball was chosen at random.
1  2/7.
72/7.
= 5/7.
6 * 5 * 4 * 3 * 2 * 1 divided by 2 * 1.
So, the correct answer to me is 5/7.
Since, 2 red, 3 green, and 2 blue, are present,
Therefore,
S=7 (numbers of the balls)
Therefore,
Let E be the event space that blue ball is been chosen at random.
n(E)=2
n(s)=7 (total numbers of balls in the bag.
Pr(E)=n(E)/n(s)
=2/7.
To find the probability that none of the blue ball is drawn, 1 must be subtracted from the probability that the blue ball was chosen at random.
1  2/7.
72/7.
= 5/7.
Sampathraj said:
6 years ago
Given criteria  Need the outcome to be either Red Or Green, but not Blue.
So, at first  The Total possible combination outcomes(Events) we would get considering Red Or Green(2+3 = 5) i.e "5c2".
Sample space(All Outcomes) = (Red+Green+Blue = 7) ; "7c2
Pr(E) = n(E)/n(S).
= 5c2 / 7c2,
=((5*4)/(2*1)) / ((7*6)/(2*1)),
=(20/2) /(42/2),
=(20/42),
=10/21,
=0.48 [ As probability defines, the chances of any Event occurrence is "Greater than or equal to '0' Or Less than or equal to '1'] [>=0 <=1].
So, at first  The Total possible combination outcomes(Events) we would get considering Red Or Green(2+3 = 5) i.e "5c2".
Sample space(All Outcomes) = (Red+Green+Blue = 7) ; "7c2
Pr(E) = n(E)/n(S).
= 5c2 / 7c2,
=((5*4)/(2*1)) / ((7*6)/(2*1)),
=(20/2) /(42/2),
=(20/42),
=10/21,
=0.48 [ As probability defines, the chances of any Event occurrence is "Greater than or equal to '0' Or Less than or equal to '1'] [>=0 <=1].
Sampathraj said:
6 years ago
Given criteria  Need the outcome to be either Red Or Green, but not Blue.
So, at first  The Total poswsible combination outcomes(Events) we would get considering Red Or Green(2+3 = 5) i.e "5c2"
Sample space(All Outcomes) = (Red+Green+Blue = 7) ; "7c2
Pr(E) = n(E)/n(S).
= 5c2 / 7c2,
=((5*4)/(2*1)) / ((7*6)/(2*1))
=(20/2) / (42/2)
=(20/42)
=10/21
=0.48 [ As probability defines, the chances of any Event occurrence is "Greater than or equal to '0' Or Less than or equal to '1'] [>=0 <=1].
So, at first  The Total poswsible combination outcomes(Events) we would get considering Red Or Green(2+3 = 5) i.e "5c2"
Sample space(All Outcomes) = (Red+Green+Blue = 7) ; "7c2
Pr(E) = n(E)/n(S).
= 5c2 / 7c2,
=((5*4)/(2*1)) / ((7*6)/(2*1))
=(20/2) / (42/2)
=(20/42)
=10/21
=0.48 [ As probability defines, the chances of any Event occurrence is "Greater than or equal to '0' Or Less than or equal to '1'] [>=0 <=1].
Krish said:
1 decade ago
Because, here we should consider no ball is blue. pro.of 2 blue balls is (2c2/7c2), pro.of 1blue ball and another 1 green ball is((2c1/7c1)*(3c1/6c1))*2.
Here 2 means b+g or g+b ways will occur. In the similar manner ,pro. of 1 b and 1 red is ((2c1/7c1)*(2c1/6c1))*2.
can u please explain how 2 comes here?.""""
But the question says NO BLUE BALL..why are we considering 1 blue and 1 green ...or 1 redor 1 blue..?
Here 2 means b+g or g+b ways will occur. In the similar manner ,pro. of 1 b and 1 red is ((2c1/7c1)*(2c1/6c1))*2.
can u please explain how 2 comes here?.""""
But the question says NO BLUE BALL..why are we considering 1 blue and 1 green ...or 1 redor 1 blue..?
Tharini said:
4 years ago
There is;
2 Red.
3 Green.
2 Blue Balls.
i.e. 5 Nonblue balls and 2 Blue.
1st ball can be picked from those 5 nonblue balls so that none of them is blue.
Hence n(E)=5 & n(S)= 7.
Hence probability is 5/7.
The next ball will be picked from remaining 6 balls and 4 NonBlue balls(Since one is already picked).
Hence Probability(2nd picked ball not blue) = 4/6.
Hence,
Total probability = (5/7) * (4/6) = 10/21.
2 Red.
3 Green.
2 Blue Balls.
i.e. 5 Nonblue balls and 2 Blue.
1st ball can be picked from those 5 nonblue balls so that none of them is blue.
Hence n(E)=5 & n(S)= 7.
Hence probability is 5/7.
The next ball will be picked from remaining 6 balls and 4 NonBlue balls(Since one is already picked).
Hence Probability(2nd picked ball not blue) = 4/6.
Hence,
Total probability = (5/7) * (4/6) = 10/21.
Lyi said:
3 years ago
The probability of the first ball not being blue is 5/7 and the probability of the second ball also not being blue is 4/6 and this is so because we have already drawn one ball and now the total number of balls got reduced to 6. Now where the 4 comes from? Its because we're assuming that the first ball we've drawn is a nonblue and therefore we have one less ball than 5 to draw from now.
I hope it was helpful.
I hope it was helpful.
(8)
Sagar said:
1 decade ago
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
= 7C2 `
= (7 x 6) /(2 x 1)
= 21.
A: Event none of the balls draw is blue.
n(A): number of possible ways = 2C1*3C1+2C2*3C0+3C2*2C0
=2*3+1*1+3*1
=6+1+3
=10.
and
P(A)=n(A)/n(S)=10/21=1/2.
Let S be the sample space.
Then, n(S) = Number of ways of drawing 2 balls out of 7
= 7C2 `
= (7 x 6) /(2 x 1)
= 21.
A: Event none of the balls draw is blue.
n(A): number of possible ways = 2C1*3C1+2C2*3C0+3C2*2C0
=2*3+1*1+3*1
=6+1+3
=10.
and
P(A)=n(A)/n(S)=10/21=1/2.
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