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A box contains 2 red, 3 green and 2 blue balls. Two bails are drawn at random. What is the probability that none of the balls drawn is blue?
- a)10/21
- b)11/21
- c)2/7
- d)5/7
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A box contains 2 red, 3 green and 2 blue balls. Two bails are drawn at...
Given:
- Total number of balls = 2 red + 3 green + 2 blue = 7
- Two balls are drawn at random
To find:
- Probability that none of the balls drawn is blue
Solution:
Let's first find the total number of ways to draw 2 balls out of 7:
Total ways = 7C2 = (7 * 6) / (2 * 1) = 21
Now, let's find the number of ways to draw 2 balls such that none of them is blue:
Number of ways = 2C0 * 5C2 + 2C1 * 5C1 + 2C2 * 5C0
- 2C0 * 5C2 represents selecting 0 blue balls out of 2 and 2 green balls out of 3
- 2C1 * 5C1 represents selecting 1 blue ball out of 2 and 1 green ball out of 3 (or 1 red ball out of 2 and 1 green ball out of 3)
- 2C2 * 5C0 represents selecting 2 blue balls out of 2 and 0 green balls out of 3 (or 2 red balls out of 2 and 0 green balls out of 3)
Number of ways = (1 * 10) + (2 * 5) + (1 * 1) = 21
Therefore, the probability of drawing 2 balls such that none of them is blue = (Number of ways to draw 2 balls such that none of them is blue) / (Total ways) = 21/21 = 1
But, we need to find the probability that none of the balls drawn is blue. So, we subtract the probability of drawing 2 blue balls from 1:
Probability that none of the balls drawn is blue = 1 - (Probability of drawing 2 blue balls) = 1 - (Number of ways to draw 2 blue balls / Total ways)
Number of ways to draw 2 blue balls = 2C2 * 5C0 = 1
Probability of drawing 2 blue balls = 1/21
Therefore, Probability that none of the balls drawn is blue = 1 - (1/21) = 20/21
Hence, option A (10/21) is incorrect and the correct answer is option B (11/21).
- Total number of balls = 2 red + 3 green + 2 blue = 7
- Two balls are drawn at random
To find:
- Probability that none of the balls drawn is blue
Solution:
Let's first find the total number of ways to draw 2 balls out of 7:
Total ways = 7C2 = (7 * 6) / (2 * 1) = 21
Now, let's find the number of ways to draw 2 balls such that none of them is blue:
Number of ways = 2C0 * 5C2 + 2C1 * 5C1 + 2C2 * 5C0
- 2C0 * 5C2 represents selecting 0 blue balls out of 2 and 2 green balls out of 3
- 2C1 * 5C1 represents selecting 1 blue ball out of 2 and 1 green ball out of 3 (or 1 red ball out of 2 and 1 green ball out of 3)
- 2C2 * 5C0 represents selecting 2 blue balls out of 2 and 0 green balls out of 3 (or 2 red balls out of 2 and 0 green balls out of 3)
Number of ways = (1 * 10) + (2 * 5) + (1 * 1) = 21
Therefore, the probability of drawing 2 balls such that none of them is blue = (Number of ways to draw 2 balls such that none of them is blue) / (Total ways) = 21/21 = 1
But, we need to find the probability that none of the balls drawn is blue. So, we subtract the probability of drawing 2 blue balls from 1:
Probability that none of the balls drawn is blue = 1 - (Probability of drawing 2 blue balls) = 1 - (Number of ways to draw 2 blue balls / Total ways)
Number of ways to draw 2 blue balls = 2C2 * 5C0 = 1
Probability of drawing 2 blue balls = 1/21
Therefore, Probability that none of the balls drawn is blue = 1 - (1/21) = 20/21
Hence, option A (10/21) is incorrect and the correct answer is option B (11/21).
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Community Answer
A box contains 2 red, 3 green and 2 blue balls. Two bails are drawn at...
Given:
- Box contains 2 red, 3 green and 2 blue balls.
- Two balls are drawn at random.
To find:
- Probability that none of the balls drawn is blue.
Solution:
Total number of ways to draw 2 balls out of 7 balls = 7C2 = 21
Let A be the event of drawing a non-blue ball in the first draw and B be the event of drawing a non-blue ball in the second draw.
P(A) = (2 red balls + 3 green balls) / 7 balls = 5/7
P(B | A) = probability of drawing a non-blue ball in the second draw given that a non-blue ball was drawn in the first draw
= (1 red ball + 2 green balls) / 6 balls = 3/6 = 1/2
P(A and B) = P(A) x P(B | A)
= (5/7) x (1/2)
= 5/14
Therefore, the probability of drawing two non-blue balls = P(A and B) = 5/14
The probability of not drawing any blue ball = 1 - probability of drawing at least one blue ball
= 1 - (number of ways to draw 1 blue ball x number of ways to draw 1 non-blue ball) / total number of ways to draw 2 balls
= 1 - (2C1 x 5C1) / 21
= 1 - (10/21)
= 11/21
Answer: The probability that none of the balls drawn is blue is 11/21, which is not among the given options. Therefore, the given answer key must be incorrect.
- Box contains 2 red, 3 green and 2 blue balls.
- Two balls are drawn at random.
To find:
- Probability that none of the balls drawn is blue.
Solution:
Total number of ways to draw 2 balls out of 7 balls = 7C2 = 21
Let A be the event of drawing a non-blue ball in the first draw and B be the event of drawing a non-blue ball in the second draw.
P(A) = (2 red balls + 3 green balls) / 7 balls = 5/7
P(B | A) = probability of drawing a non-blue ball in the second draw given that a non-blue ball was drawn in the first draw
= (1 red ball + 2 green balls) / 6 balls = 3/6 = 1/2
P(A and B) = P(A) x P(B | A)
= (5/7) x (1/2)
= 5/14
Therefore, the probability of drawing two non-blue balls = P(A and B) = 5/14
The probability of not drawing any blue ball = 1 - probability of drawing at least one blue ball
= 1 - (number of ways to draw 1 blue ball x number of ways to draw 1 non-blue ball) / total number of ways to draw 2 balls
= 1 - (2C1 x 5C1) / 21
= 1 - (10/21)
= 11/21
Answer: The probability that none of the balls drawn is blue is 11/21, which is not among the given options. Therefore, the given answer key must be incorrect.
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A box contains 2 red, 3 green and 2 blue balls. Two bails are drawn at random. What is the probability that none of the balls drawn is blue?a)10/21b)11/21c)2/7d)5/7Correct answer is option 'A'. Can you explain this answer?
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