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 2 of 11.
Krish said:
3 years ago
Using conceptual method rather than going by formula gives a better feel for what is happening.
P1 = probability of 1st ball not blue = 5/7 as 5 are nonblue.
P2 = probability of 2nd ball not blue = 4/6 as 1 nonblue is already taken out.
Probability of both, not being blue = P1xP2 = 5/7(4/6) = 10/21.
P1 = probability of 1st ball not blue = 5/7 as 5 are nonblue.
P2 = probability of 2nd ball not blue = 4/6 as 1 nonblue is already taken out.
Probability of both, not being blue = P1xP2 = 5/7(4/6) = 10/21.
(6)
ABI said:
3 years ago
The answer should be 5/7 right!
Because Probability = No.of favourable outcomes/Total outcomes.
=5/7.
Since, if both balls r not blue then they have to be either red(2) or green (3).
Thus,2+3 is 5.
So, option D is correct.
Because Probability = No.of favourable outcomes/Total outcomes.
=5/7.
Since, if both balls r not blue then they have to be either red(2) or green (3).
Thus,2+3 is 5.
So, option D is correct.
Summi said:
3 years ago
How come 7*6/2*1? please clear me.
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.
HEMALATHA said:
4 years ago
@Pranav.
The answer is 62.
The answer is 62.
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.
Christiano ronaldo said:
5 years ago
I didn't understand this. Please, anyone, help me to get it.
Arnika said:
5 years ago
What is C and S please explain?
Manish Kumar Gupta said:
5 years ago
How come this 2c2/7c2?
Harry said:
5 years ago
Can someone help me with this part of the question:
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.
I don't understand what 7C2 means or how we get 7x6 or 2x1 then divided.
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.
I don't understand what 7C2 means or how we get 7x6 or 2x1 then divided.
(1)
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