Question

suppose a drawer contains seven red socks, four brown socks, and five green socks. we draw one sock from the drawer and it is equally likely that any one of the socks is drawn. find the probabilities of the events in parts (a)-(e).
a. find the probability that the sock is brown.
(type an integer or a simplified fraction.)
b. find the probability that the sock is red or green.
(type an integer or a simplified fraction.)
c. find the probability that the sock is blue.
(type an integer or a simplified fraction.)
d. find the probability that the sock is not red.
(type an integer or a simplified fraction.)
e. we reach into the drawer without looking to pull out four socks. what is the probability that we get at least two socks of the same color?
(type an integer or a simplified fraction.)

Explanation:

Step1: Calculate total number of socks

First, we find the total number of socks. There are 7 red, 4 brown, and 5 green socks. Wait, but for part c, we need to check blue socks. Wait, the problem says "seven red socks, four brown socks, and five green socks" – so no blue socks? Wait, maybe I misread. Wait, the drawer has 7 red, 4 brown, 5 green. Wait, but part c is "Find the probability that the sock is blue." But there are no blue socks? Wait, maybe the original problem has blue? Wait, no, the user's image: "seven red socks, four brown socks, and five green socks". So total socks: \(7 + 4 + 5 = 16\) (for parts a - d, since we draw one sock). Wait, part e is drawing four socks. Let's handle part a first:

Part a: Probability sock is brown

Step1: Count brown socks and total socks

Number of brown socks: 4. Total socks: \(7 + 4 + 5 = 16\).

Step2: Calculate probability

Probability = (Number of brown socks) / (Total socks) = \(\frac{4}{16} = \frac{1}{4}\).

Part b: Probability sock is red or green

Step1: Count red and green socks

Red: 7, Green: 5. So red or green: \(7 + 5 = 12\). Total socks: 16.

Step2: Calculate probability

Probability = \(\frac{12}{16} = \frac{3}{4}\).

Part c: Probability sock is blue

Step1: Count blue socks

There are 0 blue socks (since only red, brown, green). Total socks: 16.

Step2: Calculate probability

Probability = \(\frac{0}{16} = 0\).

Part d: Probability sock is not red

Step1: Count non - red socks

Total socks: 16, red: 7. So non - red: \(16 - 7 = 9\) (brown + green: \(4 + 5 = 9\)).

Step2: Calculate probability

Probability = \(\frac{9}{16}\).

Part e: Probability at least two same color (drawing 4 socks)

This is a pigeonhole principle problem. The number of colors is 3 (red, brown, green). When drawing 4 socks, by pigeonhole principle, at least two must be the same color (since 4 socks and 3 colors: \( \lceil \frac{4}{3} ceil = 2\)). Wait, but to calculate probability, we can use 1 - probability of all different colors. But since there are only 3 colors, we can't have 4 different colors. So the probability of at least two same color is 1? Wait, no, wait: the number of colors is 3 (red, brown, green). So when drawing 4 socks, the maximum number of different colors is 3 (one of each color, then the fourth must match one of the three). So the number of ways to have all different colors: 0, because we can't have 4 different colors (only 3 colors). Wait, no: wait, the colors are red, brown, green (3 colors). So when drawing 4 socks, the possible color combinations: we can have at most 3 different colors (one red, one brown, one green, and then the fourth is one of the three). So the event "all socks different colors" is impossible? Wait, no, the number of colors is 3, so we can have at most 3 distinct colors in 4 socks. So the number of ways to have all distinct colors: 0? Wait, no, let's think in terms of combinations. The total number of ways to draw 4 socks from 16: \(C(16,4)=\frac{16!}{4!(16 - 4)!}=\frac{16\times15\times14\times13}{4\times3\times2\times1}=1820\). The number of ways to have all different colors: but since there are only 3 colors, we can't have 4 different colors. So the number of ways to have all distinct colors is 0. Therefore, the probability of at least two same color is \(1-\frac{0}{1820}=1\). Wait, but that seems odd. Wait, maybe I made a mistake. Wait, the colors are red (7), brown (4), green (5). So to have all different colors in 4 socks, we need one of each color and one more, but there are only 3 colors. So the fourth sock must be one of the three colors, so it will match one of the first three. So yes, when drawing 4 socks from 3 colors, at least two must be the same color. So the probability is 1.

But let's check part by part:

Part a:

Total socks: \(7 + 4 + 5 = 16\). Brown socks: 4. Probability: \(\frac{4}{16}=\frac{1}{4}\).

Part b:

Red or green: \(7 + 5 = 12\). Probability: \(\frac{12}{16}=\frac{3}{4}\).

Part c:

Blue socks: 0. Probability: \(\frac{0}{16}=0\).

Part d:

Not red: total - red = \(16 - 7 = 9\). Probability: \(\frac{9}{16}\).

Part e:

As explained, when drawing 4 socks from 3 colors, by pigeonhole principle, at least two must be the same color. So probability is 1.

Answer:

a. \(\frac{1}{4}\)

b. \(\frac{3}{4}\)

c. \(0\)

d. \(\frac{9}{16}\)

e. \(1\)