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 5 of 12.
Kk585 said:
1 decade ago
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 none of ball is blue ---> 1-(1/21) = 20/21
Why is my answer wrong please explain?
p(a)+p(a(bar)) = 1 .......
Sathya said:
9 years ago
Here, using combination formulae is good so the answer is 11/21 is correct.
Solution:.
First possibility 3 balls 7c2 = 21.
Second possibility is 5c2 = 11.
So that the answer is=11/21.
Solution:.
First possibility 3 balls 7c2 = 21.
Second possibility is 5c2 = 11.
So that the answer is=11/21.
Anand Nithi said:
8 years ago
R2 ,G3, B2.
Total = 2+3+2=7.
probability = 2/7 *3/7 * 2/7 = 12/21.
either R or G ..but not B.
probability of B = 2/21.
So,
(12/21) - (2/21) = 10/21.
Is this correct?
Total = 2+3+2=7.
probability = 2/7 *3/7 * 2/7 = 12/21.
either R or G ..but not B.
probability of B = 2/21.
So,
(12/21) - (2/21) = 10/21.
Is this correct?
(1)
Kkp said:
8 years ago
No answer is none of these.
If same colour balls are identical.
total possibilities are rr, rg, rb, gg, gb, bb.
We don't want bb, gb, rb.
So, probability is 3/6 = 1/2.
If same colour balls are identical.
total possibilities are rr, rg, rb, gg, gb, bb.
We don't want bb, gb, rb.
So, probability is 3/6 = 1/2.
Deepali said:
8 years ago
I understand the explained method but 7C2 means factorial of 7 so there should be 7*6*5*4*3*2*1/2*1. Wwhy in explanation they took only 7*6. Please explain it.
Prabhat kumar mishra said:
1 decade ago
Total number of balls = 7.
Therefore first probability is = 5/7and second probability will be 4/6.
Hence, total probability = (5/7)*(4/6) = 20/42 = 10/21.
Therefore first probability is = 5/7and second probability will be 4/6.
Hence, total probability = (5/7)*(4/6) = 20/42 = 10/21.
Shah Rukh said:
9 years ago
@Pooja for A. White.
Ans = totql balls = 5+6+2+2 = 15.
Totql no. Of outcomes i.e. S=15c1 =15/1 = 15.
Event A = 5c1 = 5.
Probability of event A = 5/15 = 1/3.
Ans = totql balls = 5+6+2+2 = 15.
Totql no. Of outcomes i.e. S=15c1 =15/1 = 15.
Event A = 5c1 = 5.
Probability of event A = 5/15 = 1/3.
Manoj said:
9 years ago
How did you obtain 21 sample spaces in the above-mentioned solution?
Can you mention all possible combination of balls?
Ex:{ HH, TT, HT, TH } for 2coins.
Can you mention all possible combination of balls?
Ex:{ HH, TT, HT, TH } for 2coins.
Ronak Mistry said:
2 years ago
Total Balls = 7.
Num of Selected Balls are 2.
So, the Total Outcomes = 7C2.
Num of Blue Balls are Selected = 5C2.
So, the Probability = 5C2/7C2 = 10/21.
Num of Selected Balls are 2.
So, the Total Outcomes = 7C2.
Num of Blue Balls are Selected = 5C2.
So, the Probability = 5C2/7C2 = 10/21.
(16)
Anil pradhan said:
1 decade ago
Simple and clear way is:
Probability of getting blue is 2/7.
And getting non-blue is 1-2/7 = 5/7.
i.e Total probability-Probability of blue ball.
Probability of getting blue is 2/7.
And getting non-blue is 1-2/7 = 5/7.
i.e Total probability-Probability of blue ball.
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