Final answer:
The probability that the experiment of tossing a coin ends with two consecutive heads, given the probability of a head is 1/3, is calculated by summing an infinite geometric series, yielding an answer of 1/3.
Step-by-step explanation:
The question asks for the probability that an experiment of tossing a coin will end with two consecutive heads given that the probability of getting a head in a single toss is 1/3. We can model this probability using a geometric progression for the outcomes where the experiment ends with heads: HH, THH, TTHH, TTTHH, and so on.
The first H can either come immediately or after any number of Ts. Crucially, the sequence must end with HH, and each H has a probability of 1/3 while each T has a probability of 2/3.
We calculate the sum of the probabilities of these outcomes:
(1/3)(1/3) + (2/3)(1/3)(1/3) + (2/3)(2/3)(1/3)(1/3) + ... = (1/9) + (2/27) + (4/81) + ...
This is an infinite geometric series with the first term a = (1/9) and a common ratio r = (2/3).
The sum of an infinite geometric series is given by S = a/(1 - r). Substituting the values, we get:
S = (1/9) / (1 - (2/3)) = (1/9) / (1/3) = 1/3.
Therefore, the correct answer is (A) 1/3. This represents the probability that the experiment stops with a head, given that the probability of tossing a head is 1/3.