Thread: Conditional Probability (ii)

1. Conditional Probability (ii)

box A contains 2 red balls,
box B contains 2 white balls,
box C contains 1 red ball and 1 white ball,

A box is selected at random (with equal probabilities) and one ball is taken at random from that box.

(i) Compute the probability of selecting a white ball using Bayes Rule.

A(1/3) _____ Red(1/6)
.........|____ Red(1/6)

B(1/3)______White(1/6)
........|_____White(1/6)

C(1/3)______Red(1/6)
........|_____White(1/6)

so.. P(White) = [(1/3)*(1/6)] + [(1/3)*(1/6)] + [(1/3)*(1/6)]
= 1/6

Correct?

(ii)Compute the conditional probability that B has been selected given that the ball is white.

P(B|White) = P(B intersect White) / P(White)

what is my P(B intersect White)?

2. Originally Posted by nameck
box A contains 2 red balls,
box B contains 2 white balls,
box C contains 1 red ball and 1 white ball,

A box is selected at random (with equal probabilities) and one ball is taken at random from that box.

(i) Compute the probability of selecting a white ball using Bayes Rule.

A(1/3) _____ Red(1/6)
.........|____ Red(1/6)

B(1/3)______White(1/6)
........|_____White(1/6)

C(1/3)______Red(1/6)
........|_____White(1/6)

so.. P(White) = [(1/3)*(1/6)] + [(1/3)*(1/6)] + [(1/3)*(1/6)]
= 1/6

Correct?

W=event of choosing a white ball
A=selecting from box A
B=selecting from box B
C=selecting from box C

$P(W)=P(W|A)P(A)+P(W|B)P(B)+P(W|C)P(C)$.

3. Originally Posted by danneedshelp
w=event of choosing a white ball
a=selecting from box a
b=selecting from box b
c=selecting from box c

$p(w)=p(w|a)p(a)+p(w|b)p(b)+p(w|c)p(c)$.
p(w|a) = 0 ?

4. Originally Posted by nameck
p(w|a) = 0 ?
Si

5. Originally Posted by Danneedshelp
W=event of choosing a white ball
A=selecting from box A
B=selecting from box B
C=selecting from box C

$P(W)=P(W|A)P(A)+P(W|B)P(B)+P(W|C)P(C)$.
P(W|A)P(A) = 0
P(W|B)P(B) = (1/6 + 1/6)(1/3) = 2/9
P(W|C)P(C) = (1/6)(1/3) = 1/18

P(W)=P(W|A)P(A)+P(W|B)P(B)+P(W|C)P(C)
......= 0 + 2/9 + 1/18
......= 5/18

correct?

6. Originally Posted by nameck
P(W|A)P(A) = 0
P(W|B)P(B) = (1/6 + 1/6)(1/3) = 2/9
P(W|C)P(C) = (1/6)(1/3) = 1/18

P(W)=P(W|A)P(A)+P(W|B)P(B)+P(W|C)P(C)
......= 0 + 2/9 + 1/18
......= 5/18

correct?
No, P(W|A) represents the probability of selecting a white ball, given you are selecting from box A. Since there are no white balls in box A, the probability P(W|A)=0. So, following that train of thought we have

P(W|A)P(A)=(0/2)(1/3)=0
P(W|B)P(B)=(2/2)(1/3)=(1/3)
P(W|C)P(C)=(1/2)(1/3)=(1/6).

7. understood..
what bout question (ii)?

8. Originally Posted by nameck
understood..
what bout question (ii)?
The below equation is from wikipedia:Bayes' theorem - Wikipedia, the free encyclopedia

In terms of your problem, you need to compute

$P(B|W)=\frac{P(W|B)P(B)}
{P(W|A)P(A)+P(W|B)P(B)+P(W|C)P(C)}$
(we are partitioning the boxes).

Notice you have already figured out both the numerator and denominator.