Answer :
Let's break down the problem step by step.
First, we will calculate the individual probabilities for each critic to give a favorable review based on their odds.
Critic 1:
The odds are 3 to 2, which means for every 3 favorable outcomes, there are 2 unfavorable outcomes, or 3 favorable outcomes out of a total of 3 + 2 = 5 outcomes.
The probability is:
[tex]\[ P(Critic1) = \frac{Favorable}{Total} = \frac{3}{3 + 2} = \frac{3}{5} \][/tex]
Critic 2:
The odds are 4 to 3, that means for every 4 favorable outcomes, there are 3 unfavorable outcomes, or 4 favorable outcomes out of a total of 4 + 3 = 7 outcomes.
The probability is:
[tex]\[ P(Critic2) = \frac{Favorable}{Total} = \frac{4}{4 + 3} = \frac{4}{7} \][/tex]
Critic 3:
The odds are 2 to 3, which means for every 2 favorable outcomes, there are 3 unfavorable outcomes, or 2 favorable outcomes out of a total of 2 + 3 = 5 outcomes.
The probability is:
[tex]\[ P(Critic3) = \frac{Favorable}{Total} = \frac{2}{2 + 3} = \frac{2}{5} \][/tex]
Now, we want to calculate the probability that a majority (which means at least 2 out of 3) of reviews are favorable. There are several ways this can happen:
1. Critic 1 and Critic 2 are favorable, but not Critic 3:
[tex]\[ P(Critic1) \cdot P(Critic2) \cdot (1 - P(Critic3)) = \frac{3}{5} \cdot \frac{4}{7} \cdot \left(1 - \frac{2}{5}\right) \][/tex]
2. Critic 1 and Critic 3 are favorable, but not Critic 2:
[tex]\[ P(Critic1) \cdot (1 - P(Critic2)) \cdot P(Critic3) = \frac{3}{5} \cdot \left(1 - \frac{4}{7}\right) \cdot \frac{2}{5} \][/tex]
3. Critic 2 and Critic 3 are favorable, but not Critic 1:
[tex]\[ (1 - P(Critic1)) \cdot P(Critic2) \cdot P(Critic3) = \left(1 - \frac{3}{5}\right) \cdot \frac{4}{7} \cdot \frac{2}{5} \][/tex]
4. All critics are favorable:
[tex]\[ P(Critic1) \cdot P(Critic2) \cdot P(Critic3) = \frac{3}{5} \cdot \frac{4}{7} \cdot \frac{2}{5} \][/tex]
Now, we will just sum up all these probabilities to get the final probability of the majority of favorable reviews.
Let's calculate:
1. [tex]\[ \frac{3}{5} \cdot \frac{4}{7} \cdot \left(1 - \frac{2}{5}\right) = \frac{3}{5} \cdot \frac{4}{7} \cdot \frac{3}{5} = \frac{36}{175} \][/tex]
2. [tex]\[ \frac{3}{5} \cdot \left(1 - \frac{4}{7}\right) \cdot \frac{2}{5} = \frac{3}{5} \cdot \frac{3}{7} \cdot \frac{2}{5} = \frac{18}{175} \][/tex]
3. [tex]\[ \left(1 - \frac{3}{5}\right) \cdot \frac{4}{7} \cdot \frac{2}{5} = \frac{2}{5} \cdot \frac{4}{7} \cdot \frac{2}{5} = \frac{16}{175} \][/tex]
4. [tex]\[ \frac{3}{5} \cdot \frac{4}{7} \cdot \frac{2}{5} = \frac{24}{175} \][/tex]
Adding them up:
[tex]\[ \frac{36}{175} + \frac{18}{175} + \frac{16}{175} + \frac{24}{175} = \frac{94}{175} \][/tex]
So the probability of at least two critics giving a favorable review is [tex]\( \frac{94}{175} \)[/tex].
First, we will calculate the individual probabilities for each critic to give a favorable review based on their odds.
Critic 1:
The odds are 3 to 2, which means for every 3 favorable outcomes, there are 2 unfavorable outcomes, or 3 favorable outcomes out of a total of 3 + 2 = 5 outcomes.
The probability is:
[tex]\[ P(Critic1) = \frac{Favorable}{Total} = \frac{3}{3 + 2} = \frac{3}{5} \][/tex]
Critic 2:
The odds are 4 to 3, that means for every 4 favorable outcomes, there are 3 unfavorable outcomes, or 4 favorable outcomes out of a total of 4 + 3 = 7 outcomes.
The probability is:
[tex]\[ P(Critic2) = \frac{Favorable}{Total} = \frac{4}{4 + 3} = \frac{4}{7} \][/tex]
Critic 3:
The odds are 2 to 3, which means for every 2 favorable outcomes, there are 3 unfavorable outcomes, or 2 favorable outcomes out of a total of 2 + 3 = 5 outcomes.
The probability is:
[tex]\[ P(Critic3) = \frac{Favorable}{Total} = \frac{2}{2 + 3} = \frac{2}{5} \][/tex]
Now, we want to calculate the probability that a majority (which means at least 2 out of 3) of reviews are favorable. There are several ways this can happen:
1. Critic 1 and Critic 2 are favorable, but not Critic 3:
[tex]\[ P(Critic1) \cdot P(Critic2) \cdot (1 - P(Critic3)) = \frac{3}{5} \cdot \frac{4}{7} \cdot \left(1 - \frac{2}{5}\right) \][/tex]
2. Critic 1 and Critic 3 are favorable, but not Critic 2:
[tex]\[ P(Critic1) \cdot (1 - P(Critic2)) \cdot P(Critic3) = \frac{3}{5} \cdot \left(1 - \frac{4}{7}\right) \cdot \frac{2}{5} \][/tex]
3. Critic 2 and Critic 3 are favorable, but not Critic 1:
[tex]\[ (1 - P(Critic1)) \cdot P(Critic2) \cdot P(Critic3) = \left(1 - \frac{3}{5}\right) \cdot \frac{4}{7} \cdot \frac{2}{5} \][/tex]
4. All critics are favorable:
[tex]\[ P(Critic1) \cdot P(Critic2) \cdot P(Critic3) = \frac{3}{5} \cdot \frac{4}{7} \cdot \frac{2}{5} \][/tex]
Now, we will just sum up all these probabilities to get the final probability of the majority of favorable reviews.
Let's calculate:
1. [tex]\[ \frac{3}{5} \cdot \frac{4}{7} \cdot \left(1 - \frac{2}{5}\right) = \frac{3}{5} \cdot \frac{4}{7} \cdot \frac{3}{5} = \frac{36}{175} \][/tex]
2. [tex]\[ \frac{3}{5} \cdot \left(1 - \frac{4}{7}\right) \cdot \frac{2}{5} = \frac{3}{5} \cdot \frac{3}{7} \cdot \frac{2}{5} = \frac{18}{175} \][/tex]
3. [tex]\[ \left(1 - \frac{3}{5}\right) \cdot \frac{4}{7} \cdot \frac{2}{5} = \frac{2}{5} \cdot \frac{4}{7} \cdot \frac{2}{5} = \frac{16}{175} \][/tex]
4. [tex]\[ \frac{3}{5} \cdot \frac{4}{7} \cdot \frac{2}{5} = \frac{24}{175} \][/tex]
Adding them up:
[tex]\[ \frac{36}{175} + \frac{18}{175} + \frac{16}{175} + \frac{24}{175} = \frac{94}{175} \][/tex]
So the probability of at least two critics giving a favorable review is [tex]\( \frac{94}{175} \)[/tex].
