Chapter 4, Problem 2CTC

To determine

To find: The probabilities of getting at least on 6 in 4 rolls of 1 die and getting at least on double 6 in 24 rolls of 2 dice.

To explain: The reason for Chevalier de Mere won majority of the time on the first game but lost the majority of the time when playing the second game.

Answer to Problem 2CTC

The probability of getting at least one 6 is 0.518.

The probability of getting at least one double 6 is 0.4914.

The reason isthe probability of getting at least one 6 is more than the probability of getting at least one double 6.

Explanation of Solution

Given info:

A die is rolled 4 times and 2 dice rolled 24 times.

Calculation:

Multiplicationrule:

If the A and B are independent, then $P\left(A\text{ and }B\right)=P\left(A\right)P\left(B\right)$.

Probability of getting at least one 6:

Onrolling a single die has 6 different outcomes. That is, the outcomes are ‘1, 2, 3, 4, 5, and 6’. Thus, the total number of outcomes is 6.

Let event A denote that the outcome is not a 6. Hence, the possible outcomes that the number is not 6 are ‘1, 2, 3, 4 and 5’. That is, there are 5 outcomes for eventA.

The formula for probability of event A is,

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

Substitute 5 for ‘Number of outcomes inA occurs’ and 6 for ‘Total number of outcomes in the sample space’,

$P\left(A\right)=\frac{5}{\text{6}}$

Here, the single dice is rolled four times.

Then, the probabilitythatat least one 6is,

$\begin{array}{c}P\left(\text{At least one 6}\right)=1-P\left(\text{No 6}\right)\\ =1-P\left(A\text{ and }A\text{ and }A\text{ and }A\right)\\ =1-P\left(A\right)\times P\left(A\right)\times P\left(A\right)\times P\left(A\right)\\ =1-\left(\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\right)\\ =1-0.482\end{array}$

$=0.518$

Therefore, the probability of getting at least one 6is 0.518.

Probability of getting at least one double 6:

Onrolling a two die has 36 different outcomes. That is, the outcomes are

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

Thus, the total number of outcomes is 36.

Let event B denote that the outcome is not a double 6. Hence, the possible outcomes that the number is not double 6 areexcept $\left(6,6\right)$. That is, there are 35 outcomes for eventA.

The formula for probability of event B is,

$P\left(B\right)=\frac{\text{Number of outcomes in }B}{\text{Total number of outcomes in the sample space}}$

Substitute 35 for ‘Number of outcomes inB occurs’ and 36 for ‘Total number of outcomes in the sample space’,

$P\left(B\right)=\frac{35}{\text{36}}$

Here, the two dice is rolled 24 times.

Then, the probabilitythat at least one double 6 is,

$\begin{array}{c}P\left(\text{At least one double 6}\right)=1-P\left(\text{No double 6}\right)\\ =1-{\left\{P\left(B\right)\right\}}^{24}\\ =1-{\left(\frac{35}{36}\right)}^{24}\\ =1-0.5086\end{array}$

$=0.4914$

Therefore, the probability of getting at least one double 6 is 0.4914.

From the above results it can be observed that Therefore, the probability of getting at least one 6 is more than the probability of getting at least one double 6.

Hence, Chevalier de Mere won majority of the time on the first game but lost the majority of the time when playing the second game.