Question

One of two urns is chosen at random with one as likely to be chosen as the other. Then a ball is drawn from the chosen urn. Urn 1 contains 1 white and 3 red balls, and Urn 2 has 2 whites, and 2 red balls. If a white ball is drawn what is the probability that it came from urn 2?

Answer 1

please help me i dont understand

Answer 2

// conditional probability
P(urn 2|white ball) = P(from urn 2 AND white ball)/P(white ball)

Answer 3

@Anita505 you got this?

Answer 4

no can u please expand

Answer 5

put \(A\) as the event you pick from urn 2
\(B\) as the event you get a white ball.
you want
\[P(A|B)\] so you understand that notation ?

Answer 6

*DO you understand that notation is what i meant

Answer 7

how do i plug in the numbers into that notation?

Answer 8

you do not. you have to know that
\[P(A|B)=\frac{P(A\cap B)}{P(B)}\]

Answer 9

ok then how do i find the answer from that?

Answer 10

so you need both \(P(A\cap B)\) and \(P(B)\)

Answer 11

is the answer 2?

Answer 12

you can't have a probability greater than 1

Answer 13

hmm i see that you are somewhat confused
2 is not the answer to any probability question

Answer 14

the easy part is computing \(P(A\cap B)\)
finding \(P(B)\) is a bit harder

Answer 15

okay so the answer is 4 for P(AnB)

Answer 16

P(A AND B) = 2/4 right?

Answer 17

yes

Answer 18

but it will be 1/2 if u simplify it

Answer 19

yes

Answer 20

not to butt in again, but i think
\[P(A\cap B)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}\]

Answer 21

that is my answer 1/4?

Answer 22

you have to pick from urn 2 AND get a white ball

Answer 23

no, that is the numerator of
\[\frac{P(A\cap B)}{P(B)}\]

Answer 24

is the first 1/2 from the fact that you can choose either urn

Answer 25

yes

Answer 26

\(A\cap B\) means you pick from urn 2 AND get white ball, so both things have to happen

Answer 27

ok

Answer 28

so the answer is 1/4 then

Answer 29

now comes the harder part, finding \(P(B)\)

Answer 30

ok

Answer 31

how do i find P(B)
?

Answer 32

how can you pick a white ball?
1) you pick from urn 1 AND get a white ball
OR
2) you pick from urn 2 AND get a white ball

Answer 33

ok

Answer 34

P(from urn 1 AND white ball) + P(from urn 2 AND white ball)

Answer 35

the probability you pick from urn 1 and get a white ball is
\[\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}\]the probability you pick from urn 2 and get a white ball we just computed, it is \(\frac{1}{4}\)
add them up and get
\[\frac{1}{4}+\frac{1}{8}=\frac{3}{8}\]

Answer 36

your final answer is therefore
\[\frac{\frac{1}{4}}{\frac{3}{8}}=\frac{1}{4}\times \frac{8}{3}=\frac{2}{3}\]

Answer 37

2/3?