Throw a coin until a head turns up for the 2nd time, where p is the probability of a throw results in heads and we assume that the outcome of each throw is independent to the previous outcomes. Let x be the number of items we have thrown the coin. 1) Determine p(x=2), p(x=3), and p(x=4). 2) Show that p(x=n) = (n-1)p2(1-p)n-2 for n greater than and equal to 2. Help me to get a clear answer. Thankyou and God bless!
asked Dec 13, 2016 in Statistics Answers by Kiana (120 points)

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:

To avoid this verification in future, please log in or register.

1 Answer

Please read through this completely because at first you will think the answer is wrong!

For two throws the outcomes are HH, HT, TH, TT. Only one of these has two heads, probability is 1/4, assuming equal probabilities for a head and tail.

For three throws the outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Since the coin is only thrown till the second head appears we want HTH, THH only, so the probability is 2/8=1/4.

4 throws: HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT. So of these we are looking for those that end in H and only have one other H: HTTH, THTH, TTHH, probability 3/16.

5 throws: we can eliminate those that end in T and have more than 2 H's. We're left with HTTTH, THTTH, TTHTH, TTTHH. Probability: 4/32=1/8.

To generalise we can see that the number of throws, n, has 2^n outcomes. There are n-1 positions for the first H, so we have p(x=n)=(n-1)/2^n as the probability.


n=2: p(2)=1/4; n=3: p(3)=2/8=1/4; n=4: p(4)=3/16; p(5)=4/32=1/8.

This is not the answer we expect so we need to make a small adjustment. If P is the probability of throwing a head and P≠1/2 necessarily, then we need to bring in the probability of throwing a tail=1-P.

The outcomes don't change but the probability does.

For example, when n=5 we have HTTTH, THTTH, TTHTH, TTTHH.

The probability of each of these is P^2(1-P)^3 and there are 4 of them so the overall probability is 4P^2(1-P)^3.

The general case is (n-1)P^2(1-P)^(n-2). This matches the expected answer.

Now let's work out p(2)=P^2; p(3)=2P^2(1-P); p(4)=3P^2(1-P)^2.

If we put P=1/2 (an unbiased coin) then P=1-P and we arrive at the probabilities we found earlier when the premise was that the probabilities of throwing a head or tail were equal.

answered Dec 13, 2016 by Rod Top Rated User (442,580 points)
Welcome to, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
78,532 questions
82,371 answers
63,428 users