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A person is doing 3 trials in 2 slot machines A and B.

The chance of winning in machine A is 0.4 and in B is 0.2.

In the first trial, this person choose the machine at random.

In each stage, if he wins, he would again choose the same machine, else he chooses the other machine.

1) what is the probability that the 2nd trial is machine A?

2) same question, if we know that the 3rd trial is done on machine A?

solution:

1) $\mathsf P(A_2)~=~\mathsf P(A_1,A_2)+\mathsf P(B_1,A_2)~=~1/2\cdot 0.4+1/2\cdot (1-0.2)~=~0.6$

2) $\mathsf P(A_2\mid A_3)~=~[\mathsf P(A_3\mid A_2)\cdot P(A_2)]/\mathsf P(A_3)~=~[0.4\cdot 0.6]/\mathsf P(A_3)$

I need help with calculating $\mathsf P(A_3)$ pls.

tnx.

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1) what is the probability that the 2nd trial is machine A?

1) $\mathsf P(A_2)$ $=\mathsf P(A_1,A_2)+\mathsf P(B_1,A_2)\\=\tfrac 1 2\cdot 0.4+\tfrac 1 2\cdot (1-0.2)\\=0.6$

Yes, indeed it is.

Note: You have: $\mathsf P(A_2) =\mathsf P(A_2\mid A_1)\cdot \mathsf P(A_1)+\mathsf P(A_2\mid B_1)\cdot \mathsf P(B_1)$


2) same question, if we know that the 3rd trial is done on machine A?

2) $\mathsf P(A_2\mid A_3)$ $=\dfrac{\mathsf P(A_3\mid A_2)\cdot \mathsf P(A_2)]}{\mathsf P(A_3)}\\=\dfrac{[0.4*0.6]}{\mathsf P(A_3)}$

So far, so good.

By the Law of Total Probability:$$\begin{align}\mathsf P(A_3) ~ = & ~\mathsf P(A_3\mid A_2)\cdot \mathsf P(A_2)+\mathsf P(A_3\mid B_2)\cdot \mathsf P(B_2) \end{align}$$

You can complete this now.

I got 14/25, am I right?

Short answer: Yes.

Longer answer: If you did this:-

$$\begin{align}\mathsf P(A_3) ~ = & ~\mathsf P(A_3\mid A_2)\cdot \mathsf P(A_2)+\mathsf P(A_3\mid B_2)\cdot \mathsf P(B_2) \\ ~ = & ~ (0.4)\cdot (0.6)+(1-0.2)\cdot (1-0.6) \\ ~ = & ~ 0.56 \end{align}$$

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  • $\begingroup$ I got 14/25, am I right? @Graham Kemp $\endgroup$ – Talor T Mar 15 '16 at 5:14
  • $\begingroup$ @TalorT Yes. (Added text) $\endgroup$ – Graham Kemp Mar 15 '16 at 5:25

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