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 6 of 12.
Sandeep said:
1 decade 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 = 21-11 = 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 = 21-11 = 10.
So probability of the same = 10/21 :).
Navcool said:
1 decade ago
Is it with or without replacement?
Manjunath said:
1 decade ago
In the same question what is the probability of taking 2 blue balls?
PRASHANT SINHA said:
1 decade ago
In the same question what is the probability of taking 2 blue balls?
Supriya said:
1 decade ago
How the (7*6)(2*1) has come?
Please clear my concept.
Please clear my concept.
Zuay said:
1 decade ago
No no no! I don't understand are we not suppose to consider probability of not obtaining blue balls here? then its 5/7.
Sundry said:
1 decade ago
From the explanation, where did 6 come from? why did you multiply 7 by 6? I did not understand that one.
Bhavna said:
1 decade ago
I didn't understand the method which sagar did? can anybody make it clear!
Hari said:
1 decade ago
Supriya c2 means multiply with number below it. Suppose 2c2= 2*1.
John said:
1 decade ago
Probability of occurrence of blue ball.
1 blue AND 1 blue OR,
1 red/green AND 1 blue OR,
1 blue AND 1 red/green.
(2/7 * 1/6) + (5/7 * 2/6) + (2/7 * 5/6) = 11/21.
Probability of non-occurrence of blue ball = 1-(11/21) = (10/21) ANS.
1 blue AND 1 blue OR,
1 red/green AND 1 blue OR,
1 blue AND 1 red/green.
(2/7 * 1/6) + (5/7 * 2/6) + (2/7 * 5/6) = 11/21.
Probability of non-occurrence of blue ball = 1-(11/21) = (10/21) ANS.
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