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Consider the following experiment.

Step 1. Flip a fair coin twice.

Step 2. If the outcomes are (TAILS, HEADS) then output Y and stop.

Step 3. If the outcomes are either (HEADS, HEADS) or (HEADS, TAILS), then output N and stop.

Step 4. If the outcomes are (TAILS, TAILS), then go to Step 1.

The probability that the output of the experiment is Y is (up to two decimal places)________?


Somewhere it explain as:

$$1/4 + 1/4*1/4 + 1/4*1/4*1/4 + …..$$ Approx $0.33$

Can you explain in formal way, please?

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Well,

  1. The output will be $Y$ if you get $TH$ on your first trial with chance $1/4$. So $P(Y_1) = 1/4$ or
  2. You got $TT$ on the first trial with chance $1/4$ and then repeat and get $TH$ on the second. So $P(Y_2) = (1/4)(1/4)$ or
  3. You got $TT$ on 1 and 2, and then got $TH$ on the third. So $P(Y_3) = (1/4)(1/4)(1/4)$

etc. Since the events are mutually exclusive, you can just add up the probabilities. Thus \begin{align*} P(Y) &= P(Y_1)+P(Y_2)+P(Y_3)+\dotsb\\ &= \frac{1}{4}+\left(\frac{1}{4}\right)^2+\left(\frac{1}{4}\right)^3+\dotsb \\ &= \frac{1}{4}\sum_{k=0}^\infty \left(\frac{1}{4}\right)^k \\& = \frac{1/4}{1-1/4} \\ &= \frac{1/4}{3/4} \\ &= \frac{1}{3}\\ &\approx 0.33 \end{align*}

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    $\begingroup$ @MithleshUpadhayay No problem, good luck. $\endgroup$ – Em. Feb 14 '16 at 8:02
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Probability of Y in the first run is 1/4, that on the second run is 1/4 times 1/4 etc., because you get Y in the second run when you get TT in the first run, go to step 1 and then get TH.

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Consider the different cases (i.e. after how many loops you can get a $Y$). It's easy to show that the chance of getting a $Y$ in the first try is $\frac{1}{4}$. The chance of case $3$ happening is $\frac{1}{4}$ too. Then, you repeat the loop. So, the chance that we get a $Y$ in the second loop is $\frac{1}{4} \cdot \frac{1}{4}$. Similarly, you'll se that the chance of getting a $Y$ in the $n$th step would be $\left(\frac{1}{4}\right)^n$. As the cases are exclusive and complementary, the chance of getting a $Y$ as output would be $\frac{1}{4}+\frac{1}{4} \cdot \frac{1}{4}+\frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4}...= \sum\limits_{i=1}^{\infty}{\left(\frac{1}{4}\right)^i}=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}\approx 0.33$

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    $\begingroup$ You are welcome! $\endgroup$ – SinTan1729 Feb 14 '16 at 8:27

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