# Binomial Probability Distribution: application problems HELP [closed]

I've been working on 2 problems which I really need help on: pretty sure that they are both binomial probability distribution problems. First one, I have the answer written down, but I think it is not correct. I don't really have clue on where to start with the second problem. Your help and advice will be greatly appreciated!

$$1.$$ You have $$100$$ coins, and $$99$$ of them are fair (equal probability of heads or tails). One of them is weighted and has a $$90%$$ probability of landing on heads. You randomly choose one of the $$100$$ coins. Find the probability that it is a weighted coin, under the following scenarios:

$$1$$) You flip it $$10$$ times and lands on heads $$10$$ times

$$2$$) You flip it $$10$$ times and it lands on heads $$9$$ times 1. A professor gives students a pop quiz with 5 true or false questions. Eighty percent of the students are well-prepared for the pop quiz, but twenty percent are not. Students who are prepared have a 90% chance of answering each question correctly, but the students who are unprepared simply randomly guess and have a 50% chance. Find the probability that a student was well-prepared under the following scenarios:

(a) Answered 1 correctly

## closed as off-topic by lulu, Did, Cesareo, Shailesh, Mostafa AyazSep 23 '18 at 19:07

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• These are questions involving Bayes Theorem...you don't appear to be using that. – lulu Sep 23 '18 at 10:29

$$1.1)$$ Let $$H$$ represent a coin landing on heads and $$C_W$$ represent a weighted coin being selected, then:

$$P(C_W|10H)= \frac{P(C_W\cap10H)}{P(10H)}= \frac{00.01*0.9^{10}}{0.01*0.9^{10}+0.99*0.5^{10}} \approx 0.783$$

$$1.2)$$ Using the above notation,

$$P(C_W|9H)= \frac{P(C_W\cap9H)}{P(9H)}=\frac{0.01*10C1*0.9^{9}*0.1}{0.01*10C1*0.9^{9}*0.1+0.99*10C1*0.5^{10}} \approx 0.286$$

$$2.1)$$ Let $$A$$ represent a question being correctly answered and $$W$$ represent a student who is prepared, then:

$$P(W|1A) = \frac{P(W\cap1A)}{P(1A)} = \frac{0.8*5C1*0.9*0.1^4}{0.8*5C1*0.9*0.1^4+0.2*5C1*0.5^5} \approx 0.0114$$

These questions are related to conditional probability and it would be helpful for you to read up on it.