been solving probabilities for couple of weeks now and got stuck on couple of them, this is heavily related to bayes' theorem although solvable without it. I am not interested in answers as i am in solutions, any amount of help will be appreciated
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city owns two taxi companies: "green" which owns 73 cars and "yellow" which owns 140 cars. A car has made a crash and ran away from the accident area. There was a single witness who saw a yellow car. it is known that during same circumstances witnesses give correct feedback with a probability of 0.84. what is the probability that "yellow" company driver is the one who made a crash.
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10 students out of whom 4 are "good friends", ordered cinema spots in the last row which has 10 seats. All the "good friends" except for one arrived together and got spots next to one another. what is the probability of last friend getting a seat next to his friends?
Best Answer
Bayes theorem: $$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$
For Q.1: Let A = It was a Yellow Taxi and B = The Witness Seen a Yellow Taxi $$P(A|B) = \frac{(0.84)(\frac{140}{213})}{(0.84)(\frac{140}{213})+(0.16)(\frac{73}{213})} = 0.90965$$
For Q.2: Let A = He got a seat next to his friends and B = His 3 friends got seats next to each other $$P(A|B)=\frac{(\frac{10C4}{10!})}{(\frac{9C3}{9!})} = \frac{1}{4}$$