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Chapter 1: Problem 3

A coin is to be tossed until a head appears twice in a row. What is the sample space for this experiment? If the coin is fair, what is the probability that it will be tossed exactly four times?

Short Answer

Expert verified
The sample space for tossing a coin until a head appears twice in a row consists of infinite possible sequences like HH, THH, TTHH, TTTHH, and so on. For a fair coin, the probability of tossing it exactly four times until two consecutive heads appear, i.e., TTHH, is \(\frac{1}{16}\).

Step by step solution

01

Identify the sample space of the experiment

To identify the sample space of tossing a coin until we get two heads in a row, we need to consider all the possible sequences of flips that result in two consecutive heads. The sequences can be of different lengths, but they must end in two consecutive heads. Here are the possible sequences: - HH - THH - TTHH - TTTHH, and so on. As we can see, for each additional 'T' in the sequence, the length of the sequence increases by 1. Since there's no upper bound on the length, the sample space is infinite.
02

Determine the probability of each outcome for a fair coin

Now that we have identified the sample space of the experiment, we need to determine the probability of each possible outcome for a fair coin. A fair coin has a probability of 1/2 for heads and 1/2 for tails. For example, the probability of the first outcome, HH, is (\(\frac{1}{2}\))^2 = \(\frac{1}{4}\). To generalize the probability for any sequence with k tails followed by two heads, we can use the following formula: P(k tails followed by two heads) = (\(\frac{1}{2}\))^(k + 2) Here, k + 2 refers to the total number of tosses in the sequence.
03

Calculate the probability of tossing the coin exactly four times

Now that we have determined the probability of each outcome for a fair coin, we need to find the probability of tossing the coin exactly four times. In this case, our sequence looks like TTHH, which has 2 tails followed by two heads. To find this probability, we can plug k = 2 in the formula from Step 2: P(2 tails followed by two heads) = (\(\frac{1}{2}\))^(2 + 2) = (\(\frac{1}{2}\))^4 = \(\frac{1}{16}\) Thus, the probability of tossing a fair coin exactly four times until two consecutive heads appear is \(\frac{1}{16}\).

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Most popular questions from this chapter

Three prisoners are informed by their jailer that one of them has been chosen at random to be executed, and the other two are to be freed. Prisoner \(A\) asks the jailer to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information, since he already knows that at least one will go free. The jailer refuses to answer this question, pointing out that if \(A\) knew which of his fellows were to be set free, then his own probability of being executed would rise from \(\frac{1}{3}\) to \(\frac{1}{2}\), since he would then be one of two prisoners. What do you think of the jailer's reasoning?

Suppose we have ten coins which are such that if the \(i\) th one is flipped then heads will appear with probability \(i / 10, i=1,2, \ldots, 10\). When one of the coins is randomly selected and flipped, it shows heads. What is the conditional probability that it was the fifth coin?

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Show that $$ P\left(\bigcup_{i=1}^{n} E_{i}\right) \leqslant \sum_{i=1}^{n} P\left(E_{i}\right) $$ This is known as Boole's inequality. Hint: Either use Equation (1.2) and mathematical induction, or else show that \(\bigcup_{i=1}^{n} E_{i}=\bigcup_{i=1}^{n} F_{i}\), where \(F_{1}=E_{1}, F_{i}=E_{i} \bigcap_{j=1}^{i-1} E_{j}^{c}\), and use property (iii) of a probability.

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