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\(3-12\) . Find the expected value (or expectation) of the games described. A bag contains eight white balls and two black balls. John picks two balls at random from the bag, and he wins \(\$ 5\) if he does not pick a black ball.

Short Answer

Expert verified
The expected value of the game is approximately \$3.11.

Step by step solution

01

Understanding the Problem

We are tasked with finding the expected value of a game where John wins $5 if he picks two white balls from a bag containing 8 white balls and 2 black balls.
02

Calculate Total Possible Outcomes

John picks two balls from a total of 10 balls (8 white + 2 black). The total number of ways to pick 2 balls from 10 is given by the combination formula: \( \binom{10}{2} \). Thus, \( \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \).
03

Calculate Favorable Outcomes (No Black Balls)

John picks two white balls. The number of ways to pick 2 white balls out of 8 is given by \( \binom{8}{2} \). Thus, \( \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 \).
04

Determine Probability of Winning

The probability of picking two white balls is the number of favorable outcomes divided by the total possible outcomes: \( P(\text{two white}) = \frac{28}{45} \).
05

Calculate the Expected Value of the Game

The expected value (E) is calculated by multiplying the amount won (\\(5) by the probability of winning: \( E = 5 \times \frac{28}{45} = \frac{140}{45} \approx 3.11 \). Thus, the expected value of the game is approximately \\)3.11.

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Key Concepts

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

Probability
Probability is fundamental when dealing with uncertainties and random events, like drawing balls at random from a bag. In this game, probability helps determine John's chances of winning.
The core principle of probability tells us how likely an event is to happen out of all possible outcomes. For example, when John draws two balls from the bag, we want to know the likelihood that both will be white.
To calculate probability, use:
  • Identify the total number of outcomes.
  • Count how many outcomes meet the criteria (favorable outcomes).
  • Divide the number of favorable outcomes by the total outcomes.
For John's game, the probability of drawing two white balls (since this is the winning condition) is \(\frac{28}{45}\). This is calculated by dividing the number of ways to draw two white balls by the total ways to draw any two balls from the bag.
Getting comfortable with probability helps in predicting chances and making decisions based on available data.
Combinatorics
Combinatorics is all about counting possibilities. In the context of this exercise, it provides the tools needed to count how many ways John can draw balls from the bag.
The essential formula used here is the combination formula, which is represented as \( \binom{n}{k} \). It tells us how many ways we can choose \( k \) items from \( n \) items without regard to order.
In John's situation:
  • To find total outcomes when picking 2 balls from 10, we use: \( \binom{10}{2} = 45 \).
  • To find favorable outcomes for picking 2 white balls from 8 white, we have: \( \binom{8}{2} = 28 \).

Combinatorics simplifies complex problems by providing a structured way to break them down, making it easier to handle large sets of options.
White Balls
In the context of this game, the focus is on the white balls because winning relies on whether John picks them.
The bag initially contains eight white balls, which are the targets to win the prize.
The two balls John picks have to be white for him to receive the $5 reward.
Why focus on white balls? Because only drawing white balls results in a win. To understand this:
  • The probability of drawing white balls indicates winning chances.
  • The counting of outcomes and calculation of probabilities revolve around ensuring those outcomes involve no black balls.

By understanding and focusing on the desired outcomes, he can realize the conditions needed for victory, ensuring he calculates that correctly to know his true chances.

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Most popular questions from this chapter

Smokers and Nonsmokers The participants at a mathematics conference are housed dormitory-style, five to a room. Because of an oversight, conference organizers forgot to ask whether the participants are smokers. In fact, it turns out that 30\(\%\) are smokers. Find the probability that Fred, a nonsmoking conference participant, will be housed with (a) Exactly one smoker (b) One or more smokers

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\(3-12\) . Find the expected value (or expectation) of the games described. Tim wins \(\$ 3\) if a coin toss shows heads or \(\$ 2\) if it shows tails.

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