(a) A loaded die has probabilities$\frac{\mathbf{1}}{\mathbf{21}}\mathbf{,}\frac{\mathbf{2}}{\mathbf{21}}\mathbf{,}\frac{\mathbf{3}}{\mathbf{21}}\mathbf{,}\frac{\mathbf{4}}{\mathbf{21}}\mathbf{,}\frac{\mathbf{5}}{\mathbf{21}}\mathbf{,}\frac{\mathbf{6}}{\mathbf{21}}$of showing 1, 2, 3, 4, 5, 6.What is the probability of throwing two 3’s in succession?

(b) What is the probability of throwing a 4 the first time and not a 4 the second

Time with a die loaded as in (a)?

(c) If two dice loaded as in (a) are thrown, and we know that the sum of the

numbers on the faces is greater than or equal to 10, what is the probability

That both are 5s?

(d) How many times must we throw a die loaded as in (a) to have probability greater than $\frac{\mathbf{1}}{\mathbf{2}}$of getting an ace?

(e) Adie, loaded as in (a), is thrown twice. What is the probability that thenumber on the die is even the first timethe second time?

A loaded dice which has a probability of getting 1, 2, 3, 4, 5 and 6 as$\frac{1}{21},\frac{2}{21},\frac{3}{21},\frac{4}{21},\frac{5}{21},\frac{6}{21}.$.

The events are said to be independent when the occurrence or non-occurrence of any event does not have any effect on the occurrence or non-occurrence of the other event.

When events are independent, apply the formula $P\left(AB\right)=P\left(A\right)·P\left(B\right)$where A and B are the events.

When two die are rolled, there are 36 points in the sample space.

Find the sample space for the given problem.

$\begin{array}{cccccc}1,1& 1,2& 1,3& 1,4& 1,5& 1,6\\ 2,1& 2,2& 2,3& 2,4& 2,5& 2,6\\ 3,1& 3,2& 3,3& 3,4& 3,5& 3,6\\ 4,1& 4,2& 4,3& 4,4& 4,5& 4,6\\ 5,1& 5,2& 5,3& 5,4& 5,5& 5,6\\ 6,1& 6,2& 6,3& 6,4& 6,5& 6,6\end{array}$

When two 3s are to come in succession, the probability will be $\frac{3}{21}\times \frac{3}{21}-\frac{1}{49}$

Thus the required probability is $\frac{1}{49}$.

The probability of throwing a 4 is $\frac{4}{21}$and not throwing a 4 is $1-\frac{4}{21}=\frac{17}{21}$.

Find the probability of throwing a 4 the first time and not a 4 the secondtime.

$\frac{4}{21}\times \frac{17}{21}-\frac{68}{441}$.

Thus the required probability is $\frac{68}{441}$.

It can be observed that the point in the sample space that has sum greater than or equal to 10 are $\left(4,6\right),\left(6,4\right),\left(5,5\right),\left(5,6\right),\left(6,5\right)$.

Find the probability that the sum of the numbers on the faces is greater than or equal to 10.

$\frac{4}{21}\times \frac{6}{21}+\frac{6}{21}\times \frac{5}{21}\times \frac{5}{21}+\frac{5}{21}+\frac{6}{21}\times \frac{5}{21}+\frac{6}{21}\times \frac{6}{21}=\frac{169}{441}$

Find the probability that both numbers are 5.

$\frac{5}{21}\times \frac{5}{21}=\frac{25}{441}$

Find the probability that both are 5’s when the sum of the numbers on the faces is greater than or equal to 10 using the formula $P_A\left(B\right)=\frac{P\left(AB\right)}{P\left(A\right)}$.

$$\begin{gathered} P\left(5,5\right)=\frac{{\displaystyle \frac{25}{441}}}{169} \\ =\frac{\begin{array}{c}441\\ 25\end{array}}{169} \end{gathered}$$

Thus the required probability is $\frac{25}{169}$.

The probability of not getting a one on the die is$\frac{20}{21}$. Similarly, the probability of not getting n-ones is ${\left(\frac{20}{21}\right)}^n$.

This implies that the probability of getting a one at least one time is $1-{\left(\frac{20}{21}\right)}^n$.

To have the probability more than half, $1-{\left(\frac{20}{21}\right)}^n>\frac{1}{2}$.

Solve the obtained inequality to obtain the value of n.

$$\begin{gathered} 1-{\left(\frac{20}{21}\right)}^n>\frac{1}{2} \\ \frac{1}{}>{\left(\frac{20}{21}\right)}^n \\ In\left(\frac{1}{2}\right)>n\ln \left(\frac{20}{21}\right) \\ n<\frac{In\left({\displaystyle \frac{1}{2}}\right)}{In\left({\displaystyle \frac{20}{21}}\right)} \\ <14.203 \end{gathered}$$

The obtained value is not a whole number; thus we can assume that the required number of throws will be 15.

It can be observed that the point in the sample space that has sum greater than or equal to 10 are $\left(2,5\right),\left(2,6\right),\left(4,5\right),\left(4,6\right),\left(6,4\right),\left(6,6\right)$.

Find the probability that the number on the die is even the first time, greater than 4 in the second time

$$\begin{gathered} \frac{2}{21}\times \frac{5}{21}+\frac{2}{21}\times \frac{6}{21}+\frac{4}{21}\times \frac{5}{21}+\frac{4}{21}\times \frac{6}{21}+\frac{5}{21}+\frac{6}{21}\times \frac{6}{21}=\frac{132}{441} \\ =\frac{44}{147} \end{gathered}$$

Thus the required probability is $\frac{44}{147}$.