2.6k views

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) ____________.

edited | 2.6k views

Answer is ${0.33}$

$1^{st}$ time it is ${0.25}\left(\dfrac{1}{4}\right),$ when tail tail comes, entire process gets repeated, so next time probability of $Y$ to happen is ${0.25}\times {0.25}\left(\dfrac{1}{4}\times \dfrac{1}{4}\right),$ likewise it goes on as infinite GP

Sum of infinite GP $= \dfrac{a}{(1-r)}$

here, $a= \dfrac{1}{4}$ and $r =\dfrac {1}{4}$

so answer becomes $\dfrac{1}{3}$ i.e ${0.33}$
edited
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@Sreyas, nice explanation,but can someone explain why the following solution isn't the answer..
q=1-p(not getting Y)=3/4

so,all together....put it as....pq+p^2q+p^3q+...... +p^nq =1/4
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Why the failure of getting something other than $Y$ in the second flip is not considered for computing the probability of getting Y in next flip.
+3
let $P$ be the probability that $Y$ is printed.

one-fourth of the time we have $Y$ is printed and $\dfrac{1}{4}^{th}$ of the time the experiment is repeated all again and we have the same chance for printing $Y$ again. which can be written as,

$P=\dfrac{1}{4}+\dfrac{1}{4} P$

$\dfrac{3}{4} P = \dfrac{1}{4}$

$P=\dfrac{1}{3} = .33$

$P(TH)=\dfrac{1}{4}$

$P(HH + HT)=\dfrac{1}{2}$

now if $TT$ comes then toss again,

So, $P(TTTH)=\dfrac{1}{16}$ and so on.... $P(TH+TTTH+\ldots) = \dfrac{1}{4} + \dfrac{1}{16}+\ldots= \dfrac{1}{3}$

edited

P(getting the output Y) = P(TH) +P(TTTH) + P(TTTTTH) + P(TTTTTTTH) +  P(TTTTTTTTTH)+  P(TTTTTTTTTTTH)...........

P(getting the output Y) = 1/4 + 1/42 +1/43 +1/44 +1/45 +1/46+................ = (1/4) / ( 1- 1/4) =1/3

The correct answer is 0.33 .