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A bag contains 5 white, 7 red and 4 black balls. Four balls are drawn one by one with replacement. The chance that atleast two balls are black is
- a)243/256
- b)67/256
- c)54/256
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A bag contains 5 white, 7 red and 4 black balls. Four balls are drawn ...
Let A, B, C, D denote the events of not getting a white ball in first, second, third and fourth draw respectively.
Since the balls are drawn with replacement, therefore, A, B, C, D are independent events such that P (A) = P (B) = P (C) = P (D).
Since out of 16 balls, 11 are not white, therefore, P (A) = 11/16
∴ Required probability = P (A) . P (B) . P (C) . P (D)
=> (11/16) x (11/16) x (11/16) x (11/16) = (11/16)^4.
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Community Answer
A bag contains 5 white, 7 red and 4 black balls. Four balls are drawn ...
Problem Analysis:
We need to find the probability that at least two balls drawn are black. We can find this probability by using complementary probability, i.e., the probability that none or only one ball is black. We can use the formula for probability with replacement:
P(A) = (Number of ways A can occur) / (Total number of outcomes)
We can also use the multiplication rule of probability, which states that the probability of two or more independent events occurring together is the product of their individual probabilities.
Solution:
To find the probability that none or only one ball is black, we need to consider the following cases:
Case 1: No black ball is drawn
The total number of ways to draw four balls is 16C4 = 1820
The number of ways to draw four balls without any black ball is 11C4 = 330
Therefore, the probability of not drawing any black ball is:
P(No black ball) = 330/1820 = 33/182
Case 2: Only one black ball is drawn
The total number of ways to draw four balls is 16C4 = 1820
The number of ways to draw one black ball and three non-black balls is:
(4C1 * 12C3) = 704
Therefore, the probability of drawing only one black ball is:
P(One black ball) = 704/1820 = 176/455
Therefore, the probability of drawing at least two black balls is:
P(At least two black balls) = 1 - P(No black ball) - P(One black ball)
= 1 - 33/182 - 176/455
= 67/256
Therefore, the correct option is (b) 67/256
Final Answer:
The probability that atleast two balls are black is 67/256.
We need to find the probability that at least two balls drawn are black. We can find this probability by using complementary probability, i.e., the probability that none or only one ball is black. We can use the formula for probability with replacement:
P(A) = (Number of ways A can occur) / (Total number of outcomes)
We can also use the multiplication rule of probability, which states that the probability of two or more independent events occurring together is the product of their individual probabilities.
Solution:
To find the probability that none or only one ball is black, we need to consider the following cases:
Case 1: No black ball is drawn
The total number of ways to draw four balls is 16C4 = 1820
The number of ways to draw four balls without any black ball is 11C4 = 330
Therefore, the probability of not drawing any black ball is:
P(No black ball) = 330/1820 = 33/182
Case 2: Only one black ball is drawn
The total number of ways to draw four balls is 16C4 = 1820
The number of ways to draw one black ball and three non-black balls is:
(4C1 * 12C3) = 704
Therefore, the probability of drawing only one black ball is:
P(One black ball) = 704/1820 = 176/455
Therefore, the probability of drawing at least two black balls is:
P(At least two black balls) = 1 - P(No black ball) - P(One black ball)
= 1 - 33/182 - 176/455
= 67/256
Therefore, the correct option is (b) 67/256
Final Answer:
The probability that atleast two balls are black is 67/256.
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