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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) ____________.

edited | 1.8k views

1st time it is 0.25( 1/4 ), when tail tail comes, entire process gets repeated, so next time probability of Y to happen is 0.25*0.25 ( (1/4)*(1/4) ), likewise it goes on as infinite GP

Sum of infinite GP = a/(1-r)

here a= 1/4 and r = 1/4

so answer becomes 1/3 i.e 0.33
answered by Active (1.7k points) 5 13 29
selected by
@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
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.
Answer should be 0.33

P(TH)=1/4

P(HH + HT)=1/2

now if TT comes then toss again,

So, P(TTTH)= 1/16 . and so on.... P(TH+TTTH+........) = 1/4 + 1/16+... = 1/3
answered by Junior (759 points) 2 4 16
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 .
answered by Boss (8.8k points) 3 8 12