Chapter 13, Problem 16CT

To determine

To calculate: The probability to getting exactly $2$ fours if roll a dice five times.

Answer to Problem 16CT

Solution:

The probability to getting exactly $2$ fours if roll a dice five times is $0.1608$.

Explanation of Solution

Given information:

A die roll five times.

Formula used:

The formula to find probability for any event $E$ is;

$P\left(E\right)=\frac{\text{Number}\text{of}\text{ways}\text{that}\text{E}\text{can}\text{occur}}{\text{Number}\text{of}\text{possible}\text{outcomes}}$.

The number of ways of selecting $r$ objects from $n$ distinct objects, $r\le n$, in which repetition of objects is not allowed and order is not important is given by the formula,

$C\left(n,r\right)=\frac{n!}{\left(n-r\right)!r!}$

Calculation:

To calculate the probability to getting exactly $2$ fours if roll a dice five times, first find number of possible outcomes when roll a die five times.

There are 6 possible outcomes when roll a dice one time, so using multiplication principle;

Number of outcomes when roll a dice five times is $6\cdot 6\cdot 6\cdot 6\cdot 6=7,776$.

To determine total number of ways to get exactly 2 fours, calculate number of ways to position of 2 fours in 5 and number of ways to select remaining 3 numbers.

Since order is not important in selection so, use combinations to find number of ways to position 2 fours in 5.

Thus, number of ways to position of 2 fours in 5 is $C\left(5,2\right)=\frac{5!}{\left(5-2\right)!2!}$.

Remaining three position can filled with any of the remaining five numbers from the die.

Thus, using multiplication principle, number of ways to filled remaining three positions is $5\cdot 5\cdot 5=125$.

Therefore, number of ways to get exactly 2 fours is $C\left(5,2\right)\cdot 125=\frac{5!}{\left(5-2\right)!2!}\cdot 125$.

$\Rightarrow C\left(5,2\right)\cdot 125=\frac{5!}{3!2!}\cdot 125$.

By writing $5!$ in factorial notation,

$\Rightarrow \frac{5!}{3!2!}\cdot 125=\frac{5\cdot 4\cdot 3!}{3!2!}\cdot 125$.

$\Rightarrow \frac{5\cdot 4\cdot 3!}{3!2!}\cdot 125=\frac{5\cdot 4}{2\cdot 1}125=1250$

Hence, $\text{The probability to getting exactly fours}=\frac{1250}{7776}=0.1608$.

Therefore, the probability to getting exactly $2$ fours if roll a dice five times is $0.1608$.