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.
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)
Sampathraj said:
8 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:
8 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].
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.
Sisis said:
8 years ago
The Probability of picking 1st ball is 5/7 and second is 4/6 final probability is the product 20/42=10/21.
Precious said:
8 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.
7-2/7.
= 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.
7-2/7.
= 5/7.
Laura said:
8 years ago
How 7x6?
where did you get the 6 from?
Please explain it in detail.
where did you get the 6 from?
Please explain it in detail.
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.
Abic said:
9 years ago
The probability of getting blue is 2/7.
If we sub 1-2/7 the answer is 5/7 why it is inaccurate? Please explain me.
If we sub 1-2/7 the answer is 5/7 why it is inaccurate? Please explain me.
ASHUTOSH said:
9 years ago
Why not the answer is 5/7?
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