GATE CSE 2016 | Set 1 | Question: 29

Question

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)

Comments

For Expectation part :

E = (2 * 1/4) + (2 * 1/4) + (2 * 1/4) + 1/4*(2+E)

E= 8/3.

similar Question : 

A network router attempts to transmit a data packet. In any given attempt, the packet is successfully delivered with probability $0.7$, encounters a temporary collision and must be retransmitted (restarts the process) with probability $0.2$, or is permanently dropped due to a buffer overflow with probability $0.1$. What is the probability that the packet is successfully delivered?what is the expected number of total transmission attempts made until the packet either succeeds or drops permanently

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through tree diagram : 

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If we get (T,H) in first round then stop , so  P(Y) =1/4 .       OR    If we get (T,T) , then we have to start again from beginning. i.e.  P(Y) = (1/4)*P(Y)     

So Overall  P(Y)  = (1/4)   OR    (1/4)*P(Y)     

P(Y)  = (1/4) + (1/4)*P(Y)

Answer 1

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 ${0.33}$

Comments

@JAINchiNMay First time it is $1/4$, 2^nd time it is $1/4*1/4$ and so on. Sir has just written $0.25 = 1/4$.

That’s why $a= 1/4 \ and  \ r= 1/4$

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ohh my bad thanks

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just to point out that it is a sum of decreasing G.P

Answer 2

Answer should be 0.33

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

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

now if $TT$ comes then toss again,

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

Answer 3

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 .

Comments

I also applied the same approach... Easy to understand.

Answer 4

You can correct me if I am wrong.

Comments

bro you adding p(TH) case in the first if we get (TH) we simply stop as mentioned in question the how its possible please explain if am wroong

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@gurichouhan Yes we do stop after getting TH , the question asks what is the total probability of getting TH , we can find that by summing up probabilities of getting TH at each level of the recursion tree.

Answer 5

flipCoinTwice(){
    if (Tails, Heads)
        return Y;
    if (Heads, Heads) or (Heads, Tails)
        return N
    if (Tails, Tails)
        return flipCointTwice()
}

 

$P = Y +  flipCointTwice() \\\ Y +  flipCointTwice() \\\ flipCointTwice() \\\ Y + flipCointTwice() \\\ flipCointTwice() \\\ flipCointTwice() \\\ Y + . . . . . . $

 

consider flipCointTwice()  as a variable F

$P(F) = \frac{1}{4}$

 

$P = Y + F Y + F F Y + F F F Y + F F F F Y + . . . . .  $

 

$P = Y + F [ Y + F Y + F  F Y + F F F Y + . . . . . ] $

 

$P = Y + F [ P] $

 

$P = \frac{1}{4} + \frac{1}{4} [ P] $

 

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

 

$P = \frac{1}{3} = 0.33$

Answer 6

Here is the answer : 

Comments

Thank you