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A suitcase contains 4 blue shirts, 5 red shirts and 3 white shirts.If 4 shirts are drawn what is the probability that exactly one of them is blue ?
- a)113/218
- b)14/167
- c)123/234
- d)224/495
- e)None of these
Correct answer is option 'D'. Can you explain this answer?
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A suitcase contains 4 blue shirts, 5 red shirts and 3 white shirts.If ...
P = 4C1*8C3/12C4 = 4*56/495 = 224/495
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A suitcase contains 4 blue shirts, 5 red shirts and 3 white shirts.If ...
Question category: Quant
Given:
A suitcase contains 4 blue shirts, 5 red shirts, and 3 white shirts.
4 shirts are drawn.
To find:
The probability that exactly one of them is blue.
Solution:
Total number of shirts in the suitcase = 4 + 5 + 3 = 12
Total number of ways to draw 4 shirts out of 12 = 12C4 (combination of 4 shirts out of 12)
Number of ways to draw exactly one blue shirt and three other shirts (not blue) = 4C1 * 8C3 (combination of 1 blue shirt out of 4 and 3 shirts out of 8 which are not blue)
Therefore, the required probability = (4C1 * 8C3) / 12C4
= (4*56) / 495
= 224/495
Hence, option D is the correct answer.
Note: The formula used here is the formula for probability of an event happening exactly k times in n independent trials, which is given by nCk * p^k * (1-p)^(n-k), where p is the probability of the event happening in one trial. In this case, p = 4/12 = 1/3 (the probability of drawing a blue shirt in one trial), n = 4 (the number of trials), and k = 1 (the number of times we want the event to happen). So, the required probability is 4C1 * (1/3)^1 * (2/3)^3 = 4/9 * 8/27 = 32/81. However, this is not one of the given options, so we need to simplify our answer.
Given:
A suitcase contains 4 blue shirts, 5 red shirts, and 3 white shirts.
4 shirts are drawn.
To find:
The probability that exactly one of them is blue.
Solution:
Total number of shirts in the suitcase = 4 + 5 + 3 = 12
Total number of ways to draw 4 shirts out of 12 = 12C4 (combination of 4 shirts out of 12)
Number of ways to draw exactly one blue shirt and three other shirts (not blue) = 4C1 * 8C3 (combination of 1 blue shirt out of 4 and 3 shirts out of 8 which are not blue)
Therefore, the required probability = (4C1 * 8C3) / 12C4
= (4*56) / 495
= 224/495
Hence, option D is the correct answer.
Note: The formula used here is the formula for probability of an event happening exactly k times in n independent trials, which is given by nCk * p^k * (1-p)^(n-k), where p is the probability of the event happening in one trial. In this case, p = 4/12 = 1/3 (the probability of drawing a blue shirt in one trial), n = 4 (the number of trials), and k = 1 (the number of times we want the event to happen). So, the required probability is 4C1 * (1/3)^1 * (2/3)^3 = 4/9 * 8/27 = 32/81. However, this is not one of the given options, so we need to simplify our answer.
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