probabilities, bayes' theorem exercises

Question

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

1. 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.

2. 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?

Answer 1

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 $$P(A|B)=\frac{(\frac{10C4}{10!})}{(\frac{9C3}{9!})} = \frac{1}{4}$$

Answer 2

Using Bayes Theorem?

seems to have similar problem as such i adapted it to mine

we got 73 green cars and 140 yellow cars

73*0.16+140*0.84=129.28 we get total amount of yellow cars identified by witnesses(some of whom are incorrect)

140*0.84 = 117.6 which describes the amount of actual yellow cars

129.28/117.6= 0.90965 which is the answer.

Answer 3

7 friends are going to the cinema. They will be sitting in a row with 7 seats. What is the probability that John and Mary don't sit together?

i used this as an example to make out solution. we don't seem to care about the lineup of "good friends" and we can just take them as one person, so what we have are 6 spots taken by students, 1 taken by "good friend bunch" and 1 left for our lonely lad.

so our "bunch" sits at one of the ends with probability of 2/8 then our loner has 1/7 chance of sitting next to them

our "bunch" has 6/8 chance to sit not on one of the ends, then our loner has 2/7 chance to sit next to them, therefore

$\frac 28 \cdot \frac 17+\frac 68 \cdot \frac27=\frac {7}{28}=\frac 14$