Question

Question: To determine the probability of each outcome when a biased die is rolled, if rolling a $\mathbf{\text{2}}$or rolling a$\mathbf{\text{4}}$ is three times as likely as rolling each of the other four numbers on the die and it is equally likely to roll a $\mathbf{\text{2ora4}}$ .

Short answer

Answer

The probability of each outcome when a biased die is rolled, if rolling a \(2\) or rolling a \(4\) is three times as likely as rolling each of the other four numbers on the die and it is equally likely to roll a \(2\) or a \(4\) ,

\(P(1) = \frac{1}{{16}},P(2) = \frac{6}{{16}},P(3) = \frac{1}{{16}},P(4) = \frac{6}{{16}},P(5) = \frac{1}{{16}},P(6) = \frac{1}{{16}}{\rm{. }}\)

Step-by-step solution

Given data

Given that, when a biased die is rolled, if rolling a $2$ or rolling a $4$ is three times as likely as rolling each of the other four numbers on the die.

Concept used law of total probability

If there are n number of events in an experiment, then the sum of the probabilities of those $\mathit{n}$events is always equal to $\mathbf{1}$.

$\mathit{P}\left(A_1\right)\mathbf{+}\mathit{P}\left(A_2\right)\mathbf{+}\mathit{P}\left(A_3\right)\mathbf{+}\mathbf{\dots }\mathit{P}\left(A_n\right)\mathbf{=}\mathbf{1}$.

Solve for probability

Consider p be the probability of getting other four numbers than, the probability of getting a $2$ or getting a $4$ is $3\times ($ Probability of getting other $4)=12p\$$ .

$P\left(2\right)+P\left(4\right)=3\times \left(P\right(1)+P(3)+P(5)+P(6\left)\right)=12p$

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 four numbers and the probability of getting a

$\text{2ora4is1}$.

$$\begin{gathered} P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)+P\left(5\right)+p\left(6\right) \\ =1p+12p+p+p+p=1 \\ =16p=1p=\frac{1}{16} \end{gathered}$$

Probability of getting other four numbers, that is, $1,3,5$ and $\text{6is}\frac{1}{16}$ and the Probability of getting a $2$ or a $\text{4is}\frac{1}{2}\times \frac{12}{16}=\frac{1}{16}$.

The probability of each outcome when a biased die is rolled, if rolling a 2 or rolling a 4 is three times as likely as rolling each of the other four numbers on the die and it is equally likely to roll a 2 or a 4 ,$\Rightarrow P\left(1\right)=\frac{1}{16},P\left(2\right)=\frac{6}{16},P\left(3\right)=\frac{1}{16},P\left(4\right)=\frac{6}{16},P\left(5\right)=\frac{1}{16},P\left(6\right)=\frac{1}{16}\text{.}$