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:
114 comments Page 1 of 12.
Abhikant said:
3 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.
(1)
Malik Ahamed said:
7 months ago
Where did 6 came from? Please explain.
(1)
Sneha said:
10 months ago
Can anyone explain this in the short method?
(6)
Riafat Hussain said:
11 months ago
1st ball = 5/7.
2nd ball = 4/6.
5/7 * 4/6 = 10/21.
2nd ball = 4/6.
5/7 * 4/6 = 10/21.
(51)
Ahmed Raafat said:
1 year 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)
SA said:
1 year ago
Choose the first ball: There are 7 balls in total (2 red, 3 green, and 2 blue) , so you have 7 choices for the first ball.
Choose the second ball: After picking the first ball, there are 6 balls left for the second ball (since you're not concerned with color at this point).
Now, multiply the number of choices for the first and second balls together to get the total number of ways to draw any two balls: 7 choices × 6 choices = 42 ways.
Finally, calculate the probability that none of the balls drawn is blue by dividing the number of ways to draw without blue balls by the total number of ways to draw any two balls:
Probability = (Number of ways to draw without blue balls) / (Total number of ways to draw any two balls).
Probability = 20 / 42.
You can simplify this fraction by dividing both the numerator and denominator by their greatest common divisor, which is 2:
Probability = (20 ÷ 2) / (42 ÷ 2).
Probability = 10 / 21.
So, the probability that none of the balls drawn is blue is 10/21.
Choose the second ball: After picking the first ball, there are 6 balls left for the second ball (since you're not concerned with color at this point).
Now, multiply the number of choices for the first and second balls together to get the total number of ways to draw any two balls: 7 choices × 6 choices = 42 ways.
Finally, calculate the probability that none of the balls drawn is blue by dividing the number of ways to draw without blue balls by the total number of ways to draw any two balls:
Probability = (Number of ways to draw without blue balls) / (Total number of ways to draw any two balls).
Probability = 20 / 42.
You can simplify this fraction by dividing both the numerator and denominator by their greatest common divisor, which is 2:
Probability = (20 ÷ 2) / (42 ÷ 2).
Probability = 10 / 21.
So, the probability that none of the balls drawn is blue is 10/21.
(20)
Anhad said:
2 years ago
@Saiteja
I also thought that 20/21 is correct, actually, you have only deducted the probability of both blue from 1, you have to also deduct at least one blue from 1 i.e. 10/21. So the final calculation will go 1 - 1/21 - 10/21 = 10/21. (Probability of not getting blue = 1 - the probability of both blue - probability of one blue)
I also thought that 20/21 is correct, actually, you have only deducted the probability of both blue from 1, you have to also deduct at least one blue from 1 i.e. 10/21. So the final calculation will go 1 - 1/21 - 10/21 = 10/21. (Probability of not getting blue = 1 - the probability of both blue - probability of one blue)
(12)
Saiteja said:
2 years ago
A bag contains 2 red, 3 green and 2 blue balls, total = 7 balls.
(probability of choosing+probability of not choosing=1).
Let's consider the probability of choosing two balls is blue then,
probability of choosing P=(2/7)*(1/6) = 1/21.
so, the probability of not choosing (NP) = 1-P,
NP = 1-(1/21)
NP = 20/21,
I think 20/21 is the right answer.
(probability of choosing+probability of not choosing=1).
Let's consider the probability of choosing two balls is blue then,
probability of choosing P=(2/7)*(1/6) = 1/21.
so, the probability of not choosing (NP) = 1-P,
NP = 1-(1/21)
NP = 20/21,
I think 20/21 is the right answer.
(25)
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.
(14)
Tejeshwari said:
3 years ago
5/7 is it right?
Anyone can explain it, please.
Anyone can explain it, please.
(9)
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