Chapter 13, Problem 35SGA

To determine

To find: The each pair of events is independent or dependent. And hence find the probability.

Answer to Problem 35SGA

The probability is $\frac{5}{1296}$ .

Explanation of Solution

Given:

The given events: rolling two number cubes, and getting a sum of $2$ on the first toss and a sum of $6$ on the second toss.

Calculation:

Let $A$ be the event of getting sum of $2$ on the first toss of two number cubes and $B$ be the event of getting sum of $6$ on the second toss of two number cubes.

These two events are independent events, because one event does not affect the happening of the other.

The possible outcomes after rolling two number cubes are as:

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

Possible outcomes of getting sum of $2$ on the first toss of two number cubes as:

  $\left\{\left(1,1\right)\right\}$ .

Possible outcomes of getting sum of $6$ on the first toss of two number cubes as:

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

The probability of getting sum of $2$ and probability of getting a sum of $6$ are calculated as:

  $\begin{array}{c}P\left(\text{of getting sum of}2\right)=\frac{\text{Possible number of outcomes}}{\text{Total number of outcomes}}\\ =\frac{1}{36}\\ P\left(\text{of getting sum of}6\right)=\frac{\text{Possible number of outcomes}}{\text{Total number of outcomes}}\\ =\frac{5}{36}\end{array}$

Because the above two events are independent, so

  $\begin{array}{c}P\left(A\text{and}B\right)=P\left(A\right).P\left(B\right)\\ =\frac{1}{36}.\frac{5}{36}\\ =\frac{5}{1296}\end{array}$

Therefore the probability is $\frac{5}{1296}$ .