Modification of GATE question

Question

A coin tossed 3 times

1) if output at  last two toss is $HT$ then stop tossing

2) if output of last two toss  is $TT$ then toss again and go to step 1 or step 3

3) if output of last two toss is $TH$ or $HH$ then toss again and goto step 2

What will be probability, that at last the coin will stop tossing?

Comments

Srestha I am not getting the step 2 and step 3 can u please explain it with an example.

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I mean the toss will be HHT or THT it will be stop tossing

case 2) if toss are  HTT or TTT , then it will toss again

case 3) if toss is TTH,HTH,HHH,THH , then toss again

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Only case 1 here is terminating condition
if case 2 and 3 happens, it needs to repeat toss again

Answer 1

Let us assume E and E' be two events where

E = { (HHT) , (THT) }

and E' = { (HTT), (TTT), (HTH), (TTH), (HHH), (THH) }

Then P(E) = 2/8 = 1/4  and  P(E') = 6/8 = 3/4

Now, according to the given condition in the problem at each chance we are tossing the coin 3 times , if the event E occurs then we will not get another chance to toss  and if event E' occurs then we will get another chance to toss.

Now, the probability that I got only 1 chance to toss is 1/4 ( the event E occurred on the 1st chance)

The probability that I got 2 chances to toss is 3/4 * 1/4 ( the event E' occurred on the 1st chance and event E occurred on the 2nd chance)

So, if we go on like this we can find the probability that I got k chances to toss is (3/4)^k * (1/4) (where the event E' occurred on first k-1 times after that event E occurred). 

So, the probability that tossing would terminate within k chances is

1/4 + 1/4 * 3/4 + 1/4 * (3/4)² + ..................... + 1/4 * (3/4)^k

Now, in the problem it is not given that at most how many chances we can take. So, the number of chances taken by me can run up to infinity.

So, the probability that tossing would terminate is

1/4 + 1/4 * 3/4 + 1/4 * (3/4)² + ...........................(Up to infinity)

Now, the above expression is the sum of infinite G.P series

We know that the sum of an infinite G.P. series is a/ (1-r) where a= the first term of infinite G.P. and r = common ratio

So, the sum of the above expression is (1/4) / (1- 3/4 ) = 1

So, the probability that the tossing would terminate is 1.

Hence our tossing will definitely terminate it cannot go forever.

Comments

we get HT in 1 of 4 cases

but how r u confirming it is last 2 cases?

means it is HHT and not HTT?

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Srestha I am not getting what u are asking

Extremely sorry for that. Can u please explain it a bit more elaborately.

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I mean 3 times we need to toss

and last 2 tosses are HT

now how do u know , it will be last two tosses but not 1st two tosses?

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Srestha the event E consists of two outputs one is (HHT) and another is (THT)

Now, event E occurs means one of the above two outputs occur.

Now, the above two outputs have the last two tosses as H and T

So, if the above event E occurs then the last two tosses be H and T.

But if the event E' occurs then we cannot have the last two tosses as H and T because none of the outputs in the event E' contain last two tosses as H and T.

Answer 2

i think it should be..

Comments

I think those 4 cases are not required

because , u take HHHH, which already in under HHH

So, why go upto 4 cases?

will not those are just repeating some cases?

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Because ma'am we are deciding tossing on the basis of only last two events,how will u decide a coin is tossed again or not and from where u will start ??

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but those are just repeating already occured cases
isnot it?