EM_METest

Question

I have a query here. For the case 4 when 4 tosses are made shouldn’t the probability be $\frac{1}{8}$ instead of $\frac{1}{16}$ because at 4^th toss, it be might be HHHT, or HHHH(he makes atmost 4 tosses)?

Please guide.

Comments

It'll be $1/16$ because in the first three tosses, he has to get a head and in the fourth, he has to get a tail. Both these events have a probability of 0.5 each. This is similar to Bernoulli trials with $p = 0.5$.

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Ayush HHHH is not a favorable condition.

HHHH will lead to one more toss.

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@Mk Utkarsh-It might be possible that he makes atmost 4 tosses irrespective of whether tails comes at $4^{th}$ toss or not.

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@Ayush Upadhyaya He has to make four tosses at most AND he has to get one tail. That's how you'd interpret this question.