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Fantom [35]
3 years ago
9

Your roommate Dave eats one of two things for breakfast each day. Eighty percent of the time he has eggs for breakfast, and twen

ty percent of the time he has stale bread. After eating his breakfast he walks to class and for each person he sees, he randomly chooses to either wave hello nicely or scowl angrily. If he eats eggs then each time he sees a person he has a 90% probability of waving hello nicely, and each time he sees a person he has a 10% probability of scowling angrily. If he eats stale bread, then each time he sees a person he has a 60% probability of waving hello nicely, and each time he sees a person he has a 40% probability of scowling angrily. Dave's Greeting probability for each person by breakfast choice Breakfast choice Greeting Eggs Stale bread Wave nicely 90% for each person he sees 10% Scowl angrily 60% 40% (a) Dave eats breakfast and walks to class. On his way he sees n people. Derive an expression in terms of n and k for the probability that he will scowl angrily at k people and wave nicely to n k people. In other words find the following conditional probability: P(Scowls at k|Sees n people) (b) Suppose Dave sees n people on his way to class, waves nicely to n-k and scowls at k. Derive an expression for the probability he ate stale bread for breakfast. In other words, find the following conditional probability: P(Stale bread Saw n people, scowled at k)
Mathematics
1 answer:
Usimov [2.4K]3 years ago
5 0

Answer:

9000 pounds 100%

Step-by-step explanation:

dave is fat

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<span>From the message you sent me:

when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths

If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

b_n=0.12\times b_{n-1}

Why does this work? Initially, you start with 500 mL of air that you breathe in, so b_1=500\text{ mL}. After the second breath, you have 12% of the original air left in your lungs, or b_2=0.12\timesb_1=0.12\times500=60\text{ mL}. After the third breath, you have b_3=0.12\timesb_2=0.12\times60=7.2\text{ mL}, and so on.

You can find the amount of original air left in your lungs after n breaths by solving for b_n explicitly. This isn't too hard:

b_n=0.12b_{n-1}=0.12(0.12b_{n-2})=0.12^2b_{n-2}=0.12(0.12b_{n-3})=0.12^3b_{n-3}=\cdots

and so on. The pattern is such that you arrive at

b_n=0.12^{n-1}b_1

and so the amount of air remaining after 50 breaths is

b_{50}=0.12^{50-1}b_1=0.12^{49}\times500\approx3.7918\times10^{-43}

which is a very small number close to zero.</span>
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