Chapter 13.6, Problem 29HP

To determine

Find the probability that the outcome of atleast two of the dice will be less than or equal to 4.

Answer to Problem 29HP

  $\frac{20}{27}$

Explanation of Solution

Given:

You roll 3 dice .

Formula Used:

  $P(A\text{ and }B)=P(A)\cdot P(B)$ , where A and B are independent evens

  $P(A\text{ or }B)=P(A)+P(B)$ , where A and B are independent evens

Calculation:

Rolling 3 dice , all are independent events.

The total number of outcomes = $6^3=216$ .

To find the favourable outcomes of atleast two of the dice will be less than or equal to 4:

Two dice have numbers less than or equal to 4 , or all three dice have numbers less than or equal to 4.

The favourable outcomes for getting number less than or equal to 4 on exactly two dice will be , first we choose any 2 dice from 3 dice , then in two dice , the favourable outcomes will be {1,2,3,4} and in the 3^rd dice , the favourable outcomes will be {5,6}, since it cannot have number less than or equal to 4.

  $\begin{array}{l}P(exactly\text{ two dice have number }\le \text{4)}\\ =P(only\text{ 1st two dice have number }\le 4\text{)+}P(only\text{ 1st and 3rd dice have number }\le 4\text{)+}P(only\text{ last two dice have number }\le 4\text{)}\\ =\left(\frac{4}{6}\right)\left(\frac{4}{6}\right)\left(\frac{2}{6}\right)+\left(\frac{4}{6}\right)\left(\frac{2}{6}\right)\left(\frac{4}{6}\right)+\left(\frac{2}{6}\right)\left(\frac{4}{6}\right)\left(\frac{4}{6}\right)\\ =3\cdot \frac{4}{27}=\frac{4}{9}\end{array}$

  $P(\text{exactly three dice have number }\le \text{4)=}\left(\frac{4}{6}\right)\left(\frac{4}{6}\right)\left(\frac{4}{6}\right)=\frac{8}{27}$

  $\begin{array}{l}P(\text{atleast two dice have number }\le \text{4)}\\ \text{=}P(\text{exactly two dice have number }\le \text{4)+}P(\text{exactly three dice have number }\le \text{4)}\\ \text{=}\frac{4}{9}+\frac{8}{27}\\ =\frac{12+8}{27}\\ =\frac{20}{27}\end{array}$