Answer to Question #91983 in Statistics and Probability for joe

Question

A number x is selected at random in the interval [-1, 2]. Let the events A={x<0}, B={|x-0.5|<0.5}, and C=x>0.75. Find P[A|B], P[B|C], P[A|C^c] and P[B|C^c].

Answer 1

Assume that the probability of any subinterval I of [-1, 2] is proportional to its length

$P(I)=k \cdot length(I)$

If we let $I=[-2, 1],$ then we must have

$1=P(S)=P([-2, 1])=k\cdot length([-2,1])=k\cdot 3$ $k={1 \over 3}$ $P[A]={1 \over 3}length([-1,0))={1 \over 3}\cdot 1={1 \over 3}$

$|x-0.5|<0.5=>-0.5<x-0.5<0.5=>0<x<1$

$P[B]={1 \over 3}length((0,1))={1 \over 3}\cdot 1={1 \over 3}$

$P[C]={1 \over 3}length((0.75,2))={1 \over 3}\cdot 1.25={5 \over 12}$

$P[A|B]={P[A\cap B] \over P[B]}=0$

$P[B\cap C]={1 \over 3}length((0.75,1))={1 \over 3}\cdot 0.25={1 \over 12}$

$P[B|C]={P[B\cap C] \over P[C]}={{1 \over 12} \over {5 \over 12}}={1 \over 5}$

$P[C^C]=1-{5 \over 12}={7 \over 12}$

$P[A\cap C^C]={1 \over 3}length([-1,0))={1 \over 3}\cdot 1={1 \over 3}$

$P[A|C^C]={P[A\cap C^C] \over P[C^C]}={{1 \over 3} \over {7 \over 12}}={4 \over 7}$

$P[B\cap C^C]={1 \over 3}length(0,0.75])={1 \over 3}\cdot 0.75={1 \over 4}$

$P[B|C^C]={P[B\cap C^C] \over P[C^C]}={{1 \over 4} \over {7 \over 12}}={3 \over 7}$