Question

(1 point) One of two urns is chosen at random with one just as likely to be chosen as the other. Then a ball is withdrawn from the chosen urn. Urn 1 contains 4 white and 5 red balls, and urn 2 has 3 white and 4 red balls. If a white ball is drawn, what is the probability that it came from urn 1? (Hint: Draw a tree diagram first)

Answer 1

1.Answer = 0.509

Explanation

4 white 5 Red & white 4 red Uon 2 Urn / Let a {a white ball is drawny {ith un is choseny i=1,2 . one > > Since urn is chosen out of the two ㅗ uns PCB,) = 1 P (By) = 2 Probability of white ball chosen from uon 1 is P(A/B.) a = Ho Probability of white ball chosen from uona is P(A, ) = mojte is white We have to find P(uon is chosen/ball drawn le P(B/A) By Baye's theorem, P(B/A) PCBI) . P(A/B) PCB)- P(A/B) + PCB,). P(A/B, Y₂ * 4/9 0.509 Y2x 479) + (Y2 * 37) -

2.Answer = 0.069

Explanation

In a random sample of 1000 people 6.9% have liver ailment. No; of people having tiver ailment = 69 No. of people A do not having liver ailment 931 Let £, : Person has liver ailment Fa : Person does not has liver ailment P(E) 69 1000 = .069 PCE2) = 937 = .93/ 1000 = 8% • 08 Let, A : Person is a heary drinker P(A/E) P(A/F, ) = Probability that a person selected is a heavy drinker and has liver ailment. 8% = .08 PCAlka) PC Alka) Probability that peason selected is a heavy drinker and has no liver ailment.

We have to find the probability that if a person is chosen at random and he or she is a heavy drinker, then what is the empinical probability of that person having a liver ailment. lie we have to find PCE/A). By Baye's theorem, PLE,/A) PCE). P(A/F,) PCE). P(Ale,) + P(E2). P(A/E) = 0.069 X 0.08 (.069 x 0.08) + (-93) * 0.08) = 0.069