P(A)=2/3 P(B)=1/2 P(B?A)=1/3 P(A?B)=?

[mdenham2]

P(A)=2/3 P(B)=1/2 P(B?A)=1/3 P(A?B)=?

 

[arthur ohlsten]

I believe the answer is:
P[a]=2/3
P=1/2
_
P[a]= 1/3
_
P=1/2

the following figure should help
_
-------------------------B----------------B
----------------A---1/2 x 2/3-------2/3x1/2
_
----------------A---1/2x1/3---------1/3x1/2

Probability of A given B= [1/2x1/3] /{ 1/2x2/3+ 1/2x1/3}
P[A/B]= [1/3]/[1/3+1/6]
P[A/B]=[1/3] /[ 3/6]
P[A/B]= 2/3

Arthur

 

[arthur ohlsten]

the 2nd column is B bar, and the second line is A bar. The _ symbol did not remain over the letters B and A

...............................B................B bar
....................A..........1/3..............1/3
....................A bar...... 1/6.............1/6

 

[soroban]

Hello, mdenham2!

\(\displaystyle \text{We're expected to know Bayes' Theorem: }\;P(A|B) \;=\;\frac{P(A\wedge B)}{P(B)}\)

\(\displaystyle P(A)\,=\,\frac{2}{3}\qquad P(B)\,=\,\frac{1}{2}\qquad P(B|A)\,=\,\frac{1}{3}\)

. . \(\displaystyle \text{Find: }\(A|B)\)

\(\displaystyle \text{We want: }\;P(A|B) \;=\;\frac{P(A \wedge B)}{P(B)}\) .[1]

\(\displaystyle \text{We know the denominator: }\(B) = \frac{1}{2}\)

\(\displaystyle \text{We need the numerator: }\(A\wedge B)\)


\(\displaystyle \text{We are given: }\(B|A) \,=\,\frac{1}{3}\)

. . \(\displaystyle \text{This means: }\:\frac{P(B\wedge A)}{P(A)} \,=\,\frac{1}{3}\)

. . \(\displaystyle \text{Since }P(A) = \frac{2}{3}\text{, we have: }\:\frac{P(B\wedge A)}{\frac{2}{3}} \:=\:\frac{1}{3} \quad\Rightarrow\quad P(A \wedge B) \,=\,\frac{2}{9}\)


Substitute into [1]:

. . \(\displaystyle P(A|B) \;=\;\frac{P(A\wedge B)}{P(B)} \;=\;\frac{\frac{2}{9}}{\frac{1}{2}} \;=\;\frac{4}{9}\)