Asked by W
I have been working on the following homework problem:
Consider Homer and Bart's consumption of potato chips and donuts. Suppose that Marge is in the room 30% of the time that Bart is eating chips and 40% of the time that Bart is eating donuts. Similarly, suppose that Marge is in the room 30% of the time that Homer is eating chips and 40% of the time that Homer is eating donuts; thus she catches them eating each kind of snack equally often. Nonetheless, Marge is in the room 34% of the time that Bart is eating either donuts or chips, and Marge is in the room 37% of the time that Homer is eating either donuts or chips. Suppose that Homer and Bart never eat chips and donuts in the same snack--eating chips and eating donuts are mutually exclusive.
What fraction of the time that Bart eats either chips or donuts does he eat donuts?
My work:
P(M) = probability Marge is in room
P(D) = probability Bart is eating a donut
P(C) = probability Bart is eating chips
Say P(M) = P(M|C)P(C) + P(M|D)P(D), then solving for P(D) gives P(D) = [P(M)-P(M|C)(1-P(D))]/P(M|D). P(C) is unknown but I substituted it with (1-P(D)) since we know that chip-eating and donut-eating are mutually exclusive.
Plugging in .34 for P(M), .3 for P(M|C), and .4 for P(M|D), I got P(D) = 1.4 which doesn't make any sense because probability can't be more than 1. I don't know what I am doing wrong, or if I am even in the right direction.
Any guidance would help! Thanks in advance!
Consider Homer and Bart's consumption of potato chips and donuts. Suppose that Marge is in the room 30% of the time that Bart is eating chips and 40% of the time that Bart is eating donuts. Similarly, suppose that Marge is in the room 30% of the time that Homer is eating chips and 40% of the time that Homer is eating donuts; thus she catches them eating each kind of snack equally often. Nonetheless, Marge is in the room 34% of the time that Bart is eating either donuts or chips, and Marge is in the room 37% of the time that Homer is eating either donuts or chips. Suppose that Homer and Bart never eat chips and donuts in the same snack--eating chips and eating donuts are mutually exclusive.
What fraction of the time that Bart eats either chips or donuts does he eat donuts?
My work:
P(M) = probability Marge is in room
P(D) = probability Bart is eating a donut
P(C) = probability Bart is eating chips
Say P(M) = P(M|C)P(C) + P(M|D)P(D), then solving for P(D) gives P(D) = [P(M)-P(M|C)(1-P(D))]/P(M|D). P(C) is unknown but I substituted it with (1-P(D)) since we know that chip-eating and donut-eating are mutually exclusive.
Plugging in .34 for P(M), .3 for P(M|C), and .4 for P(M|D), I got P(D) = 1.4 which doesn't make any sense because probability can't be more than 1. I don't know what I am doing wrong, or if I am even in the right direction.
Any guidance would help! Thanks in advance!
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