Question

Question: Find the probability of each outcome when a loaded die is rolled, if a 3 is twice as likely to appear as each of the other five numbers on the die.

Short answer

Answer

The probability of each outcome when a die is rolled, if a 3 is twice as likely to appear as each of the other five numbers on the die.

$$\begin{gathered} P\left(1\right)=\frac{1}{7},P\left(2\right)=\frac{1}{7}, \\ P\left(3\right)=\frac{2}{7},P\left(4\right)=\frac{1}{7}, \\ P\left(5\right)=\frac{1}{7},P\left(6\right)=\frac{1}{7}. \end{gathered}$$

Step-by-step solution

 Given  

Given that, when a die is rolled, a 3 is twice as likely to appear as each of the other five numbers on the die. Consider $p$be the probability of getting other five numbers than, the probability of getting a 3 is 2 p.

The Concept of Probability

If$\mathit{S}$represents the sample space and$\mathit{E}$represents the event. Then the probability of occurrence of favourable event is given by the formula as below:

$\mathit{P}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{E}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}$.

Determine the probability

As per the problem we have been asked to find the probability of each outcome when a die is rolled, if a 3 is twice as likely to appear as each of the other five numbers on the die.

As we know the fact that, sum of the probabilities of all the outcomes of an event is equal to 1.

So, the sum of probability of getting other five numbers and the probability of getting a 3 is 1.

$$\begin{gathered} P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)+P\left(5\right)+p\left(6\right)=1 \\ p+p+2p+p+p+p=1 \\ 7p=1 \\ p=\frac{1}{7} \end{gathered}$$

Probability of getting other five numbers, that is, 1,2,4,5 and 6 is $\frac{1}{7}$and the Probability of getting a 3 is$\frac{2}{7}$ .

The probability of each outcome when a die is rolled, if a 3 is twice as likely to appear as each of the other five numbers on the die,

$P\left(1\right)=\frac{1}{7},P\left(2\right)=\frac{1}{7},P\left(3\right)=\frac{2}{7},P\left(4\right)=\frac{1}{7},P\left(5\right)=\frac{1}{7},P\left(6\right)=\frac{1}{7}.$