Rosencrantz and Guildenstern In the opening scene of Tom Stoppard's play Rosencrantz and Guildenstern Are Dead, about two Elizabethan contemporaries of Hamlet, Guildenstern flips a coin 91 times and gets a head each time. Suppose the coin was balanced.
a. Specify the sample space for 91 coin flips, such that each outcome in the sample space is equally likely. How many outcomes are in the sample space?
b. Show Guildenstern's outcome for this sample space. Show the outcome in which only the second flip is a tail.
c. What's the probability of the event of getting a head 91 times in a row?
d. What's the probability of at least one tail, in the 91 flips?
e. State the probability model on which your solutions in parts c and d are based.
a. To specify the sample space for 91 coin flips, we need to consider the possible outcomes for each flip. Since the coin is balanced, there are two equally likely outcomes for each individual flip: heads (H) or tails (T).
For 91 flips, the sample space would consist of all possible sequences of H and T for 91 flips. Each outcome in the sample space would be a unique sequence representing a different combination of heads and tails.
The number of outcomes in the sample space can be calculated by using the formula for permutations with repetition. Since we have two possible outcomes (H and T) for each flip and we are flipping the coin 91 times, the number of outcomes in the sample space would be:
2^91 = 9.223 x 10^27
b. Guildenstern's outcome for this sample space is specified as getting a head on each of the 91 coin flips. This can be represented by a sequence of 91 H's:
H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H
The outcome in which only the second flip is a tail can be represented as:
H, T, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H, H
c. The probability of the event of getting a head 91 times in a row can be calculated by dividing the number of favorable outcomes (which is 1, as there is only one sequence with all heads) by the total number of outcomes in the sample space (2^91). Therefore, the probability would be:
Probability = 1 / (2^91)
d. The probability of getting at least one tail in the 91 flips can be calculated by subtracting the probability of getting all heads from 1. As we calculated in part c, the probability of getting all heads is 1 / (2^91). Therefore, the probability of getting at least one tail would be:
Probability of at least one tail = 1 - (1 / (2^91))
e. The solutions in parts c and d are based on the assumption that the coin is balanced, meaning that the probability of getting a head or a tail on each flip is equal. This is a Bernoulli distribution, where there are two possible outcomes with equal probabilities.