A paper bag contains a mixture of three types of treats: 12 granola bars, 10 fruit bars, and 8 cheese strips. Suppose that you play a game in which a treat is randomly taken from the bag and replaced, and then a second treat is drawn from the bag. You are allowed to keep the second treat only if it was the same type as the treat that was drawn the first time. Determine the probability that you will be able to keep a fruit bar. Use set notation in your solution.
Let's define the following events:
A: The first treat drawn is a fruit bar.
B: The second treat drawn is a fruit bar.
We want to find the probability P(B|A), which represents the probability of drawing a fruit bar on the second draw given that the first draw was a fruit bar.
To calculate P(B|A), we can use the formula:
P(B|A) = P(A ∩ B) / P(A)
Now let's calculate each term separately:
P(A): The probability of drawing a fruit bar on the first draw.
Given that the bag contains a total of 12 granola bars, 10 fruit bars, and 8 cheese strips, the probability of drawing a fruit bar on the first draw is:
P(A) = 10 / (12 + 10 + 8) = 10/30 = 1/3
P(A ∩ B): The probability of drawing a fruit bar on both draws.
Since the treats are replaced after each draw, the probability of drawing a fruit bar on both draws is simply the square of the probability of drawing a fruit bar on one draw:
P(A ∩ B) = (10/30) * (10/30) = 1/9
Now we can plug the values back into the formula for P(B|A):
P(B|A) = P(A ∩ B) / P(A) = (1/9) / (1/3) = 1/9 * 3/1 = 1/3
Therefore, the probability of being able to keep a fruit bar is 1/3, which can be written in set notation as P(B|A) = 1/3.
To determine the probability of keeping a fruit bar, we need to find the probability of drawing a fruit bar first and then drawing another fruit bar.
Let's break down the problem step by step:
Step 1: Find the probability of drawing a fruit bar first.
To do this, we divide the number of fruit bars by the total number of treats:
Probability of drawing a fruit bar = (number of fruit bars) / (total number of treats)
Probability of drawing a fruit bar = 10 / (12 + 10 + 8) = 10 / 30 = 1/3
Step 2: Find the probability of drawing another fruit bar.
Since the first treat is replaced, the probability of drawing a fruit bar again is the same as the probability of drawing a fruit bar first:
Probability of drawing another fruit bar = Probability of drawing a fruit bar = 1/3
Step 3: Find the probability of keeping a fruit bar.
Since we need to satisfy both conditions (drawing a fruit bar first and then drawing another fruit bar), we multiply the probabilities:
Probability of keeping a fruit bar = Probability of drawing a fruit bar first * Probability of drawing another fruit bar
Probability of keeping a fruit bar = (1/3) * (1/3) = 1/9
Therefore, the probability of keeping a fruit bar is 1/9, which can be represented in set notation as P(keeping a fruit bar) = 1/9.
success if you draw 2 granolas or 2 fruits or 2 cheese
prob(success) = (12/30)(12/30) + (10/30)(10/30) + (8/30)(8/30
= 4/25 + 1/9 + 16/225
= 77/225