Die A has 4 red and 2 white faces, whereas die B has

2 red and 4 white faces. A fair coin is flipped once. If it

lands on heads, the game continues with die A; if it lands on tails, then die B is to be used.

(a) Show that the probability of red at any throw is 12

(b) If the first two throws result in red, what is the probability of red at the third throw?

(c) If red turns up at the first two throws, what is the probability

that it is die A that is being used?

Given:

It is given that a bones A has 2 white and 4 red faces, and another bones B has 2 red and 4 white faces.

A show is coin is flipped formerly. .

If it lands on heads, the game continues with $\mathrm{A}$.

If it lands on tail, the game continues with B.

Calculation:

Let the events:

$A=dieA$is chosen

${\mathrm{A}}^{\mathrm{c}}=$bones $\mathrm{A}$is not chosen $=\mathrm{dieB}$is chosen

$Q_i={\mathrm{i}}^{\text{th}}$bones roll result in red

it is known that coin is fair and after flip of coin, the bones is chosen,

$P\left(A\right)=\frac{1}{2}$

$\mathrm{P}\left({\mathrm{A}}^{\mathrm{c}}\right)=\mathrm{P}\left(\mathrm{B}\right)=\frac{1}{2}$

If the number of red sides are to be noted down,

For every i,

$$\begin{gathered} P\left(\begin{array}{c}\frac{Q_i}{A}\end{array}\right)=\frac{4}{6}=\frac{2}{3} \\ P\left(\begin{array}{c}\frac{Q_i}{B}\end{array}\right)=\frac{2}{6}=\frac{1}{3} \end{gathered}$$

Now,

To calculate the probability of red at any throw,$$\begin{gathered} P\left(Q_{\mathrm{i}}\right)=\mathrm{P}\left(\frac{Q_{\mathrm{i}}}{\mathrm{A}}\right)\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\frac{Q_{\mathrm{i}}}{\mathrm{B}}\right)\mathrm{P}\left(\mathrm{B}\right) \\ =\frac{2}{3}\times \frac{1}{2}+\frac{1}{3}\times \frac{1}{2} \\ =\frac{1}{3}+\frac{1}{6} \\ =\frac{2+1}{6} \\ =\frac{3}{6} \\ =\frac{1}{2} \end{gathered}$$

Conclusion:

The probability of red at any throw is $\frac{1}{2}$.

To calculate the probability of red at the third gamble if the first two gamble result in red,

$P\left(\frac{Q_1}{Q_1{Q}_2}\right)=\frac{P\left(\frac{Q_1{Q}_2{Q}_1}{1}\right)P\left(A\right)+P\left(\frac{Q_1{Q}_2{Q}_3}{D}\right)P\left(B\right)}{P\left(\frac{Q_1{Q}_2}{A}\right)P\left(A\right)+P\left(\frac{Q_1{Q}_2}{A}\right)P\left(B\right)}$

So,

$$\begin{gathered} P\left(\frac{Q_1{Q}_2{Q}_2}{A}\right)=P\left(\frac{Q_n}{A}\right)P\left(\frac{Q_2}{A}\right)P\left(\frac{Q_2}{A}\right) \\ ={\left(\frac{2}{3}\right)}^3 \\ =\frac{B}{27} \\ P\left(\frac{Q_1{Q}_2{Q}_3}{B}\right)=P\left(\frac{Q_1}{B}\right)P\left(\frac{Q_2}{B}\right)P\left(\frac{Q_3}{B}\right) \\ ={\left(\frac{1}{3}\right)}^3 \\ P\left(\frac{Q_1{O}_2}{A}\right)=P\left(\frac{Q_1}{A}\right)P\left(\frac{Q_2}{A}\right) \\ ={\left(\frac{2}{3}\right)}^2 \\ =\frac{4}{9} \\ P\left(\frac{Q_1{O}_2}{B}\right)=P\left(\frac{Q_1}{B}\right)P\left(\frac{Q_2}{B}\right) \\ ={\left(\frac{2}{3}\right)}^2 \\ =\frac{1}{9} \end{gathered}$$

So,

$P\left(\frac{Q_h}{Q_1{Q}_2}\right)=\frac{\frac{\pi }{\pi }\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{I}}{\frac{4}{\pi }\times \frac{1}{2}+\frac{\pi }{4}\times \frac{1}{2}}$

$=\frac{\frac{3}{3}+\frac{1}{4}}{\frac{4}{4}+\frac{1}{x}}$

$=\frac{\frac{9}{3}}{\frac{3}{16}}$

$=\frac{9}{34}\times \frac{18}{5}$

$=\frac{3}{5}$

Conclusion:

The probability of red at the third gamble if the first two throws result in red is$\frac{3}{5}$

To calculate the probability that die A is used if red turns up at the first two gamble,

That is,

$P\left(\frac{A}{Q_1{O}_2}\right)$

So,

localid="1646729510868" $P\left(\frac{Q_5}{Q_1{O}_2}\right)=P\left(\frac{Q_5}{Q\left({}_1{O}_2\right.}\right)P\left(\frac{Q}{Q_1{Q}_2}\right)+P\left(\frac{Q_5}{B{O}_1{O}_2}\right)P\left(\frac{R}{Q_1{O}_2}\right)$

It is given that $Q_1,Q_2$and $Q_3$are independent given A,

It means that $P\left(\frac{O_5}{A{Q}_1{Q}_2}\right)=P\left(\frac{Q_2}{A}\right)$

So,

$$\begin{gathered} P\left(\frac{Q_1}{Q_1{O}_2}\right)=P\left(\frac{Q_4}{A}\right)P\left(\frac{Q}{Q_1{Q}_2}\right)P\left(\frac{A}{Q_1{O}_2}\right)+P\left(\frac{Q_4}{R}\right)P\left(\frac{R}{Q_1(h_2}\right) \\ \frac{3}{5}=\frac{2}{3}P\left(\begin{array}{c}\frac{A}{Q_1{O}_2}\end{array}\right)+\frac{1}{3}P\left(\begin{array}{c}\frac{R}{Q_1{Q}_2}\end{array}\right) \end{gathered}$$

So,

$$\begin{gathered} P\left(\frac{B}{Q_1{Q}_2}\right)=P\left(\frac{A^2}{Q_{142}}\right) \\ =1-P\left(\frac{A}{O_1{O}_2}\right) \end{gathered}$$

So,

$$\begin{gathered} \frac{3}{5}=\frac{2}{3}P\left(\frac{A}{Q_h{Q}_2}\right)+\frac{1}{3}P\left(\frac{B}{Q_h{Q}_2}\right) \\ \frac{3}{5}=\frac{2}{3}P\left(\frac{A}{Q_1\left(C_2\right.}\right)+\frac{1}{3}\left(1-P\left(\frac{A}{Q_1\left(h\right)}\right)\right) \\ \frac{3}{5}=\frac{1}{3}+\frac{1}{3}P\left(\frac{A}{Q_1{Q}_2}\right) \\ \frac{1}{3}P\left(\frac{A}{Q_h{Q}_2}\right)=\frac{3}{5}-\frac{1}{3} \\ \frac{1}{3}P\left(\frac{A}{Q_1{Q}_2}\right)=\frac{9-5}{15} \\ \frac{1}{3}P\left(\frac{A}{R_1{R}_2}\right)=\frac{4}{15} \\ P\left(\frac{A}{Q_1{O}_2}\right)=\frac{4}{15}\times \frac{3}{1} \\ P\left(\frac{A}{Q_1{Q}_h}\right)=\frac{4}{5} \end{gathered}$$

Conclusion:

The probability that die A is used if red turns up at the first two throws $=\frac{4}{5}$