sriharimurthy wrote:
A man chooses an outfit from 3 different shirts, 2 different pairs of shoes, and 3 different pants. If he randomly selects 1 shirt, 1 pair of shoes, and 1 pair of pants each morning for 3 days, what is the probability that he wears the same pair of shoes each day, but that no other piece of clothing is repeated?
A. \((\frac{1}{3})^6*(\frac{1}{2})^3\)
B. \((\frac{1}{3})^6*(\frac{1}{2})\)
C. \((\frac{1}{3})^4\)
D. \((\frac{1}{3})^2*(\frac{1}{2})\)
E. \(5*(\frac{1}{3})^2\)
Happy solving!
3 different shirts, 2 different pair of shoes, and 3 different pants..
For first morning, Probability of selecting 1 shirt, 1 pair of shoes, and 1 pair of pants = 3/3 * 2/2 *3/3 =1
For 2nd morning, Probability of selecting other shirt, same pair of shoes and other pair of pants = 2/3 * 1/2 * 2/3 = 2/9 (2 fresh shirts left and 2 fresh pair of pants left)
For 3rd morning, Probability of selecting other shirt, same pair of shoes and other pair of pants = 1/3 * 1/2 * 1/3 = 1/18 (1 fresh shirt left and 1 fresh pair of pants left)
So, probability that he wears the same pair of shoes each day, but that no other piece of clothing is repeated = 1 * 2/9 * 1/18 = \(\frac {1}{3^4}\)
Answer C _________________
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