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Chapter 2: Problem 55

A coin is to be tossed four times. Define events \(X\) and \(Y\) such that \(X:\) first and last coins have opposite faces \(Y\) : exactly two heads appear Assume that each of the sixteen head/tail sequences has the same probability. Evaluate (a) \(P\left(X^{C} \cap Y\right)\) (b) \(P\left(X \cap Y^{C}\right)\)

Short Answer

Expert verified
The probabilities for (a) P(X^C 鈭� Y) and (b) P(X 鈭� Y^C) are, respectively, based on the identified sequences and the calculations described in step 3.

Step by step solution

01

List Head/Tail Sequences

Start by listing all the possible sequences with 4 coin tosses. There are 16 in total. They can be represented as TTTT, TTTH, TTHT, TTHH, THTT, THTH, THHT, THHH, HTTT, HTTH, HTHT, HTHH, HTTT, HTTH, HTHT, HTHH, HTTT, HTTH, HTHT, HTHH. Here, 'H' represents 'Heads' and 'T' represents 'Tails'.
02

Identify Sequences for Events X, Y and their Complements

Next, identify sequences which meet the conditions for events X and Y and their complements. For instance, for event X, sequences with first and last coins of opposite faces would be TTHT, THTT, HTTH and HHTT, while for its complement (X^C) sequences would be TTTT, TTHH, HTTT, HTHH. Similarly, for event Y, sequences with exactly two heads would be THHT, HTHH, HHHT, HHTT, HTHH, HTHT, HHHT, while for its complement (Y^C) sequences would include TTTT, HTTT, HHHH, THHH.
03

Evaluate Probabilities

Finally, evaluate the probabilities for (a) P(X^C 鈭� Y) and (b) P(X 鈭� Y^C). The probability of any sequence is \(\frac{1}{16}\) as every sequence has the same chances of occurring. Count the number of sequences for each required event and multiply by \(\frac{1}{16}\). For example, if there are 2 sequences for P(X^C 鈭� Y), the probability will be \(2 * \frac{1}{16} = \frac{1}{8}\).
04

Calculate final values

Carrying out the calculation for both (a) P(X^C 鈭� Y) and (b) P(X 鈭� Y^C), note down the final values. This gives the required solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coin Toss Probability
Coin tossing is a classic example in probability theory, often used to understand basic probability concepts. A fair coin has two equal outcomes: heads (H) and tails (T). Thus, the probability of getting heads or tails on a single toss is \( \frac{1}{2} \).

When tossing a coin multiple times, outcomes become sequences of Hs and Ts. For example, if a coin is tossed four times, possible outcomes include sequences such as TTTT, THTH, and HTHT.

This makes the analysis of such events a study of patterns in coin tossing, which lays the foundation for understanding probability distributions and events.
Events and Outcomes
In probability, an 'event' is a specific outcome or a set of outcomes of an experiment. For instance, in the coin toss exercise, two events are defined: \( X \) where the first and last coin show opposite faces, and \( Y \) where exactly two heads appear.

An event's complement, like \( X^C \) or \( Y^C \), includes all outcomes that do not satisfy the original event. When computing probabilities, it's crucial to clearly identify these events and their complements.
  • \( X \): Opposite faces on the first and last toss mean possible sequences are like HTHT.
  • \( Y \): Exactly two heads can result in sequences such as HTHH.
This systematic identification helps in calculating probabilities for combinations such as intersections of events (e.g., \( X^C \cap Y \)).
Combinatorial Probability
Combinatorial probability deals with counting methods to evaluate the probability of an event. It involves determining all possible outcomes and the favorable ones for particular events.

In a four-coin-toss scenario, there are \( 2^4 = 16 \) possible sequences. This is because each coin has 2 outcomes (head or tail), and there are four tosses.

For complex events like \( P(X^C \cap Y) \) and \( P(X \cap Y^C) \), it's necessary to count sequences fulfilling both events' conditions. For example:
  • If 4 sequences fulfill \( X^C \cap Y \), then \( P(X^C \cap Y) = \frac{4}{16} = \frac{1}{4} \).
This approach makes it easier to visualize dependencies and relationships between events by efficiently counting outcomes.

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Most popular questions from this chapter

According to your neighborhood bookie, five horses are scheduled to run in the third race at the local track, and handicappers have assigned them the following probabilities of winning: $$ \begin{array}{lc} \hline \text { Horse } & \text { Probability of Winning } \\ \hline \text { Scorpion } & 0.10 \\ \text { Stamy Avenger } & 0.25 \\ \text { Australian Doll } & 0.15 \\ \text { Dusty Stake } & 0.30 \\ \text { Outandout } & 0.20 \\ \hline \end{array} $$ Suppose that Australian Doll and Dusty Stake are scratched from the race at the last minute. What are the chances that Outandout will prevail over the reduced field?

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A desk has three drawers. The first contains two gold coins, the second has two silver coins, and the third has one gold coin and one silver coin. A coin is drawn from a drawer selected at random. Suppose the coin selected was silver. What is the probability that the other coin in that drawer is gold?

What is the smallest number of switches wired in parallel that will give a probability of at least \(0.98\) that a circuit will be completed? Assume that each switch operates independently and will function properly \(60 \%\) of the time.
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