91Ó°ÊÓ

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,$$\begin{gathered} P\left(Q_{\mathrm{i}}\right)=\mathrm{P}\left(\frac{Q_{\mathrm{i}}}{\mathrm{A}}\right)\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\frac{Q_{\mathrm{i}}}{\mathrm{B}}\right)\mathrm{P}\left(\mathrm{B}\right) \\ =\frac{2}{3}×\frac{1}{2}+\frac{1}{3}×\frac{1}{2} \\ =\frac{1}{3}+\frac{1}{6} \\ =\frac{2+1}{6} \\ =\frac{3}{6} \\ =\frac{1}{2} \end{gathered}$$

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,

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

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

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

$=\frac{9}{34}×\frac{18}{5}$

$=\frac{3}{5}$

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,

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

Conclusion:

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