Chapter 5.1, Problem 50E

To determine

To find the probability that the sum of the dice is 5, if the red die came up 3.

Answer to Problem 50E

If the red die came up 3, the probability that the sum of the dice is 5 is $\frac{1}{6}$ and the probability is not same as it was before the red die was observed.

Explanation of Solution

Given:

One red die and one blue die are rolled. The outcome of each die is a number between 1 and 6. The ordered pairs of the outcomes are of the form (Red die, Blue die).

Calculation:

Let’s assume first value in each ordered pair denotes the Red die and second values denotes the Blue die.

If the red die comes up in 3, then blue die may comeup with 1, 2, 3, 4, 5, or 6.

Hence the sample space is,

  $\left\{\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right)\right\}$ .

Let $A$ be the event that the sum of the dice is 5, if the red die came up 3.

  $\therefore A=\left\{\left(3,2\right)\right\}$

Hence the probability that the sum of the dice is 5, if the red die came up 3 is,

  $\begin{array}{l}P\left(A\right)=\frac{\text{Number}\text{of}\text{outcomes}\text{in}A}{\text{Number}\text{of}\text{outcomes}\text{in}\text{the}\text{sample}\text{space}}\\ =\frac{1}{6}\end{array}$

Now the original sample space of the given experiment before the red die was observed is,

  $\left\{\begin{array}{l}\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right),\\ \left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right),\\ \left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right),\left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\end{array}\right\}$

Let $B$ be the event that the sum of the dice is 5.

  $\therefore B=\left\{\left(1,4\right),\left(2,3\right),\left(3,2\right),\left(4,1\right)\right\}$

So, the number of outcomes in the event $B$ is 4.

Hence, the probability that the sum of the dice is 5 is,

  $\begin{array}{l}P\left(B\right)=\frac{\text{Number}\text{of}\text{outcomes}\text{in}B}{\text{Number}\text{of}\text{outcomes}\text{in}\text{the}\text{sample}\text{space}}\\ =\frac{4}{36}\\ =\frac{1}{9}\end{array}$

Hence, if the red die came up 3, the probability that the sum of the dice is 5 is $\frac{1}{6}$ and the probability is not same as it was before the red die was observed.