Problem 10 - Algebra and Trigonometry

Question

Find the expected value (or expectation) of the games described.? $$ \begin{array}{l}{\text { A bag contains eight white balls and two black balls. John }} \\ {\text { picks two balls at random from the bag, and he wins } \$ 5 \text { if he }} \\ {\text { does not pick a black ball. }}\end{array} $$

Step-by-Step Solution

Answer

The expected value of winning is approximately $3.11.

Step 1: Understanding Probability of Winning

To win, John must not pick any black ball. There are a total of 10 balls: 8 white and 2 black. The probability of not picking a black ball in the first draw is \( \frac{8}{10} \). If the first ball is white, 9 balls remain, and the probability the second ball is also white is \( \frac{7}{9} \). Hence, the probability John wins is \( \frac{8}{10} \times \frac{7}{9} = \frac{56}{90} = \frac{28}{45} \).

Step 2: Calculate Expected Value

The expected value \( E(X) \) is calculated as the sum of all potential outcomes, each multiplied by their probability. There is only one outcome where John wins \( \$5 \) if he does not pick a black ball. The expected value is \( 5 \times \frac{28}{45} = \frac{140}{45} = \frac{28}{9} \approx 3.11 \).

Key Concepts

Probability

Probability is a fundamental concept in statistics and mathematics that measures the likelihood of an event occurring. In John's game, he is attempting to select two white balls from a bag containing eight white balls and two black balls. Our goal is to find the probability of such an event.To quantify this likelihood, we analyze the scenario of John winning the game by only picking white balls. The probability of drawing a white ball on the first attempt is calculated as the ratio of white balls to the total number of balls, i.e., \( \frac{8}{10} \). If his first pick is white, only 9 balls remain, consisting of 7 white balls and 2 black balls. Therefore, the probability of drawing another white ball becomes \( \frac{7}{9} \).These probabilities are combined through multiplication because one event follows the other. Calculating this yields the combined probability, \( \frac{8}{10} \times \frac{7}{9} = \frac{56}{90} \), which simplifies to \( \frac{28}{45} \). This value represents how likely John is to win in that scenario.

Combinatorics

Combinatorics is an area of mathematics focused on counting and arranging possibilities. It plays a vital role when analyzing probability problems like the one faced by John.In this game, combinatorics helps us determine the possible outcomes and the number of favorable outcomes. John has a bag of 10 balls, and he chooses two without any preference for order - typical of a combinatorial problem.We can use binomials to calculate the total number of ways to choose 2 balls out of 10, as described by the combination formula: \( \binom{10}{2} \). This is calculated as:\[ \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \]Similarly, by choosing only white balls, we find the counts are:Regarding only white balls, \( \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 \).Ultimately, these calculations feed into probability, revealing that out of 45 possible pairs, 28 pairs consist only of white balls. Hence, combinatorics simplifies understanding different selections, vital for computing probabilities like John's.

Random Selection

Random selection is the process of choosing items randomly without bias, meaning each item has an equal opportunity to be selected. This concept is foundational to understand John's game scenario. When John reaches into the bag to pick a ball, each of the ten balls has an equal chance of being selected on the first draw. This mechanics define the random selection process. Every draw is independent and unbiased. In practice, this means:

• Each white ball has a 1 in 10, or 10%, chance of being the first selected ball
• Once a ball is selected, the probabilities adjust based on the new group composition.
• This procedure continues until the selection process meets its criteria in this case, picking 2 balls.

Therefore, regardless of past selections, every draw from the bag remains fair and random. This unbiased approach is critical for calculating true probabilities and expected outcomes, ensuring that computations like expected value remain accurate.