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 | |||
| = 7C2 ` | ||||
|
||||
| = 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. | |||
| = 5C2 | ||||
|
||||
| = 10. |
| P(E) = |
n(E) | = | 10 | . |
| n(S) | 21 |
Discussion:
116 comments Page 3 of 12.
Kanishk said:
5 years ago
@Biswajit.
Firstly the probability of drawing two blue balls will be 2/7*1/6.
Secondly, it won't work by subtracting from 1 since two blue balls and no blue balls aren't the only two cases. There are other case balls with different colours, one of the colours being blue.
Hope you get it now!
Firstly the probability of drawing two blue balls will be 2/7*1/6.
Secondly, it won't work by subtracting from 1 since two blue balls and no blue balls aren't the only two cases. There are other case balls with different colours, one of the colours being blue.
Hope you get it now!
(3)
Shanel said:
10 years ago
Hey it should be 5/7.
Because if we subtract no of blue from total.
Then 7-(2+3) log%5 = 7*5*3*5*6 = log 7+5+3+6+1 = log 35 + log 15 + log 30.
So if we take log as common the it will be easy i.e = log (35+15+30).
= log*80 = 80 log the 80-75 = 5.
There fore the possibility is 5/7.
Because if we subtract no of blue from total.
Then 7-(2+3) log%5 = 7*5*3*5*6 = log 7+5+3+6+1 = log 35 + log 15 + log 30.
So if we take log as common the it will be easy i.e = log (35+15+30).
= log*80 = 80 log the 80-75 = 5.
There fore the possibility is 5/7.
Mohammad Javed said:
1 decade ago
There is one easy way.
No balls should be blue means either it should be red or green which sums to 5 balls. Total balls is 7.
Probability of no blue will be = (5/7)*(4/6)=10/21.
As we have picked one ball we are reducing Red+Green balls to 4. And total balls to 6.
No balls should be blue means either it should be red or green which sums to 5 balls. Total balls is 7.
Probability of no blue will be = (5/7)*(4/6)=10/21.
As we have picked one ball we are reducing Red+Green balls to 4. And total balls to 6.
Jaani! said:
1 decade ago
@krish, Cuz 2C2 means picking those 2 blue balls out of those 2 blue balls only! Bt d ques says "None of them blue" bt one blue and one of any other colour cant b included also! Thats y we do that of picking up only green and red.. (no possiblty of blue)!
Ahmed Raafat said:
2 years ago
7 CHOOSE 2 can also be denoted as 7C2.
7 is the total number of distinct elements (n),
2 is the number of elements drawn or chosen at a time (k),
21 is the total number of possible combinations (C).
nCk = n! / (k! * (n - k)!)
7C2 =7! / (2! x 5!) = 21.
7 is the total number of distinct elements (n),
2 is the number of elements drawn or chosen at a time (k),
21 is the total number of possible combinations (C).
nCk = n! / (k! * (n - k)!)
7C2 =7! / (2! x 5!) = 21.
(7)
Krish said:
1 decade ago
I am not geting wats wrong in this...please explain....
Probability that the 2 balls r blue ----> 2c2/7c2 = 1/21
Probability that none of ball is blue ---> 1-(1/21) = 20/21
Why is my answer wrong please explain?
p(a)+p(a(bar)) = 1 .......
Probability that the 2 balls r blue ----> 2c2/7c2 = 1/21
Probability that none of ball is blue ---> 1-(1/21) = 20/21
Why is my answer wrong please explain?
p(a)+p(a(bar)) = 1 .......
Sai said:
5 years ago
Total number of balls gives 2+3+2=7 s be the sample space.
number of drawing balls n(s)=2 balls out of 7.
7 c 2 = 7 * 6 = 42/2 = 21.
e = drawing balls none of which is blue,
2 balls out of 2+3=5.
5 c2.
5*4 = 20/2 = 10.
Therefore p n(e)= 10/21.
number of drawing balls n(s)=2 balls out of 7.
7 c 2 = 7 * 6 = 42/2 = 21.
e = drawing balls none of which is blue,
2 balls out of 2+3=5.
5 c2.
5*4 = 20/2 = 10.
Therefore p n(e)= 10/21.
(5)
Abhikant said:
8 months ago
Two balls removed at random,- these two balls removed one by one or at once it is not mentioned.
If it's mentioned one by one then the answer would be 10/21.
But both balls and none of the balls drawn are blue then probability = 1 - 2/7 = 5/7.
If it's mentioned one by one then the answer would be 10/21.
But both balls and none of the balls drawn are blue then probability = 1 - 2/7 = 5/7.
(10)
Akash said:
1 decade ago
Can we do it by this way?
3c2+2c2+3c1 2c1
---------------
7c2.
I have done it this way. 3C2 means that the both balls would be green 2C2 both would be red and 3C1 and 2C1 because both the balls can be red and green. Isn't the solution correct?
3c2+2c2+3c1 2c1
---------------
7c2.
I have done it this way. 3C2 means that the both balls would be green 2C2 both would be red and 3C1 and 2C1 because both the balls can be red and green. Isn't the solution correct?
Harry said:
7 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.
(2)
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