Answer to Question #232526 in Statistics and Probability for Spencer

Question

Suppose we have 3 cards identical in form except that both sides of the first card are colored red, both
sides of the second card are colored black, and one side of the third card is colored red and the other
side is colored black. The 3 cards are mixed up in a hat, and 1 card is randomly selected and put down
on the ground. If the upper side of the chosen card is colored red, what is the probability that the
other side is colored black

Answer 1

Solution.

Given the following events:

$RR=All\:side\:Red$

$BB=All\:side\:Black$

$\:RB=Red\:and\:Black\:side$

$R=Upturned\:side\:of\:the\:chosen\:card\:is\:red.$

Based on the information given in the equation we have:

$P(R|RB)=\frac{1}{2}$

$P(RB)=\frac{1}{3}$

$P(BB)=\frac{1}{3}$

$P(RR)=\frac{1}{3}$

$P(R/RR)=1$

$P(R/BB)=0$

Using the results we have to find $P(RB/R)$

Se we can say:

$P\left(RB|R\right)=\frac{P\left(RB\cap NR\right)}{P\left(R\right)}$

$=\frac{P\left(R|RB\right)P\left(RB\right)}{P\left(R|RB\right)P\left(RB\right)+P\left(R|BB\right)P\left(BB\right)P\left(R|RR\right)P\left(RR\right)}$

By substituting values we get:

$P\left(RB|R\right)=\frac{\frac{1}{2}\times \frac{1}{3}}{\left(\frac{1}{2}\times \frac{1}{3}\right)+\left(0\times \frac{1}{3}\right)+\left(1\times \frac{1}{3}\right)}=\frac{1}{3}=0.3333$

The required probability is $=\frac{1}{3}=0.3333$