Math Problem Statement

Exercise 15: A player tosses a coin n times. The probability of getting tails on a toss is p. The player wins if he gets tails exactly once. For what values of p is he most likely to win? Exercise 16: Two players of equal strength are playing against each other. What is the probability of one of the players winning 5 out of 8 games? At least 5 games out of 8?

Solution

Let's tackle each exercise one by one.

Exercise 15

The player tosses a coin nn times. The probability of getting tails on a toss is pp. The player wins if he gets tails exactly once. We need to determine the values of pp that maximize the player's chances of winning.

The probability of getting exactly one tail in nn tosses follows a binomial distribution: P(exactly 1 tail)=(n1)p(1p)n1P(\text{exactly 1 tail}) = \binom{n}{1} p (1 - p)^{n - 1} Simplifying, we have: P(exactly 1 tail)=np(1p)n1P(\text{exactly 1 tail}) = n p (1 - p)^{n - 1} To maximize this probability, we need to find the value of pp that maximizes P(exactly 1 tail)P(\text{exactly 1 tail}). This can be done by differentiating the function with respect to pp and setting it to zero.

  1. Differentiate P=np(1p)n1P = n p (1 - p)^{n - 1} with respect to pp.
  2. Solve the resulting equation to find the optimal pp.

This requires calculus, but the general solution indicates that the probability is maximized at specific values of pp, often when pp is relatively small for large nn to increase the chances of getting exactly one tail.

Exercise 16

Two players of equal strength are playing against each other. We want to find:

  1. The probability of one player winning 5 out of 8 games.
  2. The probability of one player winning at least 5 out of 8 games.

Since the players are of equal strength, each game is an independent event with a win probability of 0.5 for each player. Let XX represent the number of wins for one player out of 8 games, which follows a binomial distribution: XBinomial(8,0.5)X \sim \text{Binomial}(8, 0.5)

  1. Probability of winning exactly 5 out of 8 games: P(X=5)=(85)(0.5)5(0.5)3=(85)(0.5)8P(X = 5) = \binom{8}{5} (0.5)^5 (0.5)^{3} = \binom{8}{5} (0.5)^8

  2. Probability of winning at least 5 out of 8 games: P(X5)=P(X=5)+P(X=6)+P(X=7)+P(X=8)P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) We calculate each of these probabilities and sum them up.

Would you like a detailed solution for the differentiation in Exercise 15 or the exact calculations for Exercise 16?


Here are 5 related questions to explore further:

  1. How does the probability in Exercise 15 change if nn increases?
  2. What if the player needs to get tails exactly twice instead of once in Exercise 15?
  3. How would the probability change in Exercise 16 if one player were slightly stronger?
  4. What is the expected number of games won by a player in Exercise 16?
  5. How does the binomial distribution shape change with different values of nn and pp?

Tip: In probability problems, breaking down complex scenarios into simpler cases (like calculating for exactly 1 tail or at least 5 wins) helps make the solution manageable.

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution
Differentiation

Formulas

P(exactly 1 tail) = n * p * (1 - p)^(n - 1)
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k) for binomial probabilities

Theorems

Binomial Theorem

Suitable Grade Level

Grades 11-12