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Three players simultaneously toss coins. The coin tossed by \(A(B)[C]\) turns up heads with probability \(P_{1}\left(P_{2}\right)\left[P_{3}\right]\). If one person gets an outcome different from those of the other two, then he is the odd man out. If there is no odd man out, the players flip again and continue to do so until they get an odd man out. What is the probability that \(A\) will be the odd man?

Short Answer

Expert verified
The probability that A will be the odd man out is given by \(P_{1}(1 - P_{2})(1 - P_{3}) + (1 - P_{1})P_{2}P_{3}\).

Step by step solution

01

Case 1 - A gets heads and B, C get tails

In this case, we'll calculate the probability that A gets heads and both B and C get tails as follows: - A gets heads: \(P_{1}\) - B gets tails: \(1 - P_{2}\) - C gets tails: \(1 - P_{3}\) The probabilities need to multiply since these events are independent. So, for this case, the probability is: `Case 1 probability = \(P_{1}(1 - P_{2})(1 - P_{3})\)`.
02

Case 2 - A gets tails and B, C get heads

In this case, we'll calculate the probability that A gets tails and both B and C get heads as follows: - A gets tails: \(1 - P_{1}\) - B gets heads: \(P_{2}\) - C gets heads: \(P_{3}\) The probabilities need to multiply since these events are independent. So, for this case, the probability is: `Case 2 probability = \((1 - P_{1})P_{2}P_{3}\)`.
03

Calculate the overall probability

To find the probability that A is the odd man out, we'll consider both cases calculated in steps 1 and 2. So, we'll sum up the probabilities of both cases: `Probability(A is the odd man out) = Case 1 probability + Case 2 probability` `Probability(A is the odd man out) = \(P_{1}(1 - P_{2})(1 - P_{3}) + (1 - P_{1})P_{2}P_{3}\)`. This is the probability that A will be the odd man out in a single round of coin-tossing.

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Key Concepts

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

Independent Events
In probability theory, independent events are sets of outcomes where the occurrence of one does not affect the occurrence of others. Understanding this is crucial for situations involving multiple participants and randomness, like our coin-tossing scenario.
  • Say Player A gets heads. This event is independent from Player B's coin landing either heads or tails; similarly, Player C's result is unaffected by A's outcome.
  • Independence implies that the probability of a collection of events occurring simultaneously is the product of their individual probabilities.
For example, if A tosses a head with probability \(P_1\), B tosses a tail with \(1 - P_2\), and C tosses a tail with \(1 - P_3\), we have three independent events. Combined, the probability that these three networked events occur together is \(P_1(1 - P_2)(1 - P_3)\). This principle allows us to understand complex scenarios by dissecting them into simpler, unrelated actions.
Probability Calculation
Calculating probabilities involves combining individual probabilities, particularly when the events are independent. In our thought experiment with the players tossing coins, finding the likelihood of specific outcomes is central.In practice, we determine probabilities like this:
  • If Player A gets heads, and Player B and C both get tails, the probability is given by \(P_1(1 - P_2)(1 - P_3)\). This involves multiplying the probability of each independent event.
  • Alternatively, if Player A gets tails while Players B and C get heads, the formula is \((1 - P_1)P_2P_3\).
  • To determine the total probability of A being the odd man out, these individual probabilities are summed: \(P_1(1 - P_2)(1 - P_3) + (1 - P_1)P_2P_3\).
Breaking down complex measures into manageable parts simplifies probability calculations. Thus, understanding the roles of individual probabilities enables the solving of comprehensive issues efficiently.
Conditional Probability
Although not directly mentioned in the given exercise, understanding conditional probability enriches our grasp on independent events as it highlights scenarios where outcomes depend on others. Conditional probability might come into play if, for instance, the outcome of one player depended on another.

What Is It?

Conditional probability refers to the probability of an event occurring given another event has already occurred. It is expressed as \(P(A|B)\), meaning "the probability of A given B."

Connection To Our Exercise

However, in the context of independent events (like the coin tosses), we assume that this conditional dependence does not exist. That is, knowing the outcome of one player's coin does not change the expected probabilities for the other players.

In Other Contexts

Condition dependence adds a fascinating layer of complexity when players or events might influence each other. Yet by understanding these rules, we sharpen our ability to model various real-world scenarios through probability theory.

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Most popular questions from this chapter

The probability that a new car battery functions for over 10,000 miles is .8, the probability that it functions for over 20,000 miles is \(.4\), and the probability that it functions for over 30,000 miles is .1. If a new car battery is still working after 10,000 miles, what is the probability that (a) its total life will exceed 20,000 miles; (b) its additional life will exceed 20,000 miles?

There are two local factories that produce radios. Each radio produced at factory \(A\) is defective with probability \(.05\), whereas each one produced at factory \(B\) is defective with probability.01. Suppose you purchase two radios that were produced at the same factory, which is equally likely to have been either factory \(A\) or factory \(B\). If the first radio that you check is defective, what is the conditional probability that the other one is also defective?

Each of 2 cabinets identical in appearance has 2 drawers. Cabinet \(A\) contains a silver coin in each drawer, and cabinet \(B\) contains a silver coin in one of its drawers and a gold coin in the other. A cabinet is randomly selected, one of its drawers is opened, and a silver coin is found. What is the probability that there is a silver coin in the other drawer?

An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the um. What is the probability that all of the balls selected are white? What is the conditional probability that the die landed on 3 if all the balls selected are white?

Die \(A\) has 4 red and 2 white faces, whereas die \(B\) has 2 red and 4 white faces. A fair coin is flipped once. If it lands on heads, the game continues with die \(A\); if it lands tails, then die \(B\) is to be used. (a) Show that the probability of red at any throw is \(\frac{1}{2}\).

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