In this situation a fair coin is tossed until a head is obtained. What is the probability that the coin was tossed at least 4 times?

This question comes directly out of the 5th edition Text discrete mathematics by Kenneth Ross.

My initial intuition is that this merely requires that you get the probability that tails is obtained $3$ flips in a row which I believe is equal to $\frac{1}{8}$. My second thought however is that you also have to factor in the probability that the fourth flip is a heads. this I believe is equal to $\frac{1}{16}$.

I'm not confident however that my method for achieving these numbers is correct which is $\frac{1}{2^3}$ and $\frac{1}{2^4}$ respectively.

Thanks in advance for any guidance.

  • 1
    $\begingroup$ First thought is correct. If you start $TTT$ then you are guaranteed to require at least $4$ tosses. Conversely, any other three rolls ends your quest for $H$. $\endgroup$ – lulu May 16 '18 at 7:06

Clearly, if the coin is tossed repeatedly until the first observation of heads, then in order to toss the coin at least four times, the first three observations must be tails. This occurs with probability $1/8$, as you noted.

After the third tail, the fourth toss could be heads, in which the tossing stops, or it could be tails, in which case more trials are generated until heads is observed; but in either case, at least four coin tosses were made.

If we wanted the probability of tossing the coin exactly four times, this would be $1/16$, as there is a unique outcome that this event corresponds to: $TTTH$.


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