Question - Probability of Drawing Socks with Money
Solution:
The problem states that Ebony has 5 socks in a drawer: 4 black and 2 white. One of the white socks has a \$5 bill, one of the black socks has a \$100 bill, and another black sock has a \$40 bill.a. To list the sample points with the aid of a tree diagram, we would identify all the possible outcomes for taking two socks from the drawer. However, since I can't draw a tree diagram here, I'll describe the sample points:- BB (black/black): B1B2, B1B3, B1B4, B2B3, B2B4, B3B4 (where B1 has \$100, B2 has \$40, and B3 and B4 are without money)- WW (white/white): W1W2 (W1 has \$5, W2 has no money)- BW or WB (black/white): B1W1, B1W2, B2W1, B2W2, B3W1, B3W2, B4W1, B4W2b. The probability that Ebony chooses two socks of the same color is the sum of the probabilities of choosing two black socks (BB) plus the probability of choosing two white socks (WW). Since there are a total of 5 socks and she chooses 2, there are C(5,2) = 10 possible pairs of socks she could choose.For black socks (BB), there are C(4,2) = 6 possible pairs. For white socks (WW), there is only 1 possible pair. So the probability is calculated as follows:P(same color) = P(BB) + P(WW) = (6/10) + (1/10) = 7/10c. To calculate the probability that Ebony finds at least \$45, we need to consider the combinations of socks that include either the black sock with \$100 or the black and white socks that total at least \$45 (which can only be the black sock with \$40 and the white sock with \$5).There are two successful outcomes: (B1) and (B2W1). There are 10 possible pairs of socks she could choose. Let's calculate the probability:P(at least $45) = P(B1 or B2W1) = (Number of pairs with at least $45) / (Total number of pairs)P(at least $45) = P(B1) + P(B2W1) = (1/10) + (1/10) = 2/10 = 1/5Therefore, 1/5 is the probability that Ebony will find at least \$45.
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