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A box contains 5 Green 3 Red 4 Black balls. If two balls are drawn at random then what is the probability that there is atleast one Red?
- a)31/66
- b)13/22
- c)5/11
- d)9/11
- e)None
Correct answer is option 'C'. Can you explain this answer?
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A box contains 5 Green 3 Red 4 Black balls. If two balls are drawn at ...
(3c 1 * 9c 1 + 3c 2)/12c 2 =5/11
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A box contains 5 Green 3 Red 4 Black balls. If two balls are drawn at ...
Problem:
A box contains 5 green, 3 red, and 4 black balls. Two balls are drawn at random. Find the probability of drawing at least one red ball.
Solution:
To find the probability of drawing at least one red ball, we can calculate the probability of drawing no red balls and subtract it from 1.
Step 1: Calculate the probability of drawing no red balls
To calculate the probability of drawing no red balls, we need to consider two cases:
Case 1: Drawing two green balls
The probability of drawing a green ball on the first draw is 5/12, since there are 5 green balls out of 12 total balls.
After drawing a green ball, there are 4 green balls left out of 11 total balls. Therefore, the probability of drawing a green ball on the second draw is 4/11.
The probability of drawing two green balls is (5/12) * (4/11) = 20/132.
Case 2: Drawing one green ball and one black ball
The probability of drawing a green ball on the first draw is 5/12.
After drawing a green ball, there are 4 black balls left out of 11 total balls. Therefore, the probability of drawing a black ball on the second draw is 4/11.
The probability of drawing one green ball and one black ball is (5/12) * (4/11) = 20/132.
Step 2: Calculate the probability of drawing at least one red ball
The probability of drawing at least one red ball is 1 - (probability of drawing no red balls).
Probability of drawing no red balls = (20/132) + (20/132) = 40/132.
Therefore, the probability of drawing at least one red ball is 1 - (40/132) = 92/132 = 23/33.
Answer:
The probability of drawing at least one red ball is 23/33, which is equivalent to option C.
A box contains 5 green, 3 red, and 4 black balls. Two balls are drawn at random. Find the probability of drawing at least one red ball.
Solution:
To find the probability of drawing at least one red ball, we can calculate the probability of drawing no red balls and subtract it from 1.
Step 1: Calculate the probability of drawing no red balls
To calculate the probability of drawing no red balls, we need to consider two cases:
Case 1: Drawing two green balls
The probability of drawing a green ball on the first draw is 5/12, since there are 5 green balls out of 12 total balls.
After drawing a green ball, there are 4 green balls left out of 11 total balls. Therefore, the probability of drawing a green ball on the second draw is 4/11.
The probability of drawing two green balls is (5/12) * (4/11) = 20/132.
Case 2: Drawing one green ball and one black ball
The probability of drawing a green ball on the first draw is 5/12.
After drawing a green ball, there are 4 black balls left out of 11 total balls. Therefore, the probability of drawing a black ball on the second draw is 4/11.
The probability of drawing one green ball and one black ball is (5/12) * (4/11) = 20/132.
Step 2: Calculate the probability of drawing at least one red ball
The probability of drawing at least one red ball is 1 - (probability of drawing no red balls).
Probability of drawing no red balls = (20/132) + (20/132) = 40/132.
Therefore, the probability of drawing at least one red ball is 1 - (40/132) = 92/132 = 23/33.
Answer:
The probability of drawing at least one red ball is 23/33, which is equivalent to option C.
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