Question

Imagine performing the coin-flip experiment of Problem \(\mathrm{P} 29.19\), but instead of using a fair coin, a weighted coin is employed for which the probability of landing heads is two-fold greater than landing tails. After tossing the coin 10 times, what is the probability of observing the following specific outcomes: a. no heads b. two heads c. five heads d. eight heads

Short answer

The probabilities for each specific outcome when flipping the coin 10 times are: a. no heads: P(0) ≈ 0.000017 b. two heads: P(2) ≈ 0.00153 c. five heads: P(5) ≈ 0.1360 d. eight heads: P(8) ≈ 0.3018

Step-by-step solution

Determine the probability of landing heads and tails

Since the probability of landing heads is two-fold greater than landing tails, we can represent the probabilities as follows: P(H) = 2x P(T) = x Since the total probability is equal to 1, we can set up the equation: P(H) + P(T) = 1 Substitute the probabilities: 2x + x = 1 3x = 1 Now, solve the equation for x: x = 1/3 So, P(H) = 2/3 and P(T) = 1/3. >Find the probability of each specific outcome

Calculate the probability of no heads

Using the binomial probability formula, let x = 0 and n = 10: P(0) = \(\binom{10}{0} \cdot (2/3)^0 \cdot (1/3)^{10-0}\) P(0) = \(\binom{10}{0} \cdot 1 \cdot (1/3)^{10}\) P(0) ≈ 0.000017

Calculate the probability of two heads

Using the binomial probability formula, let x = 2 and n = 10: P(2) = \(\binom{10}{2} \cdot (2/3)^2 \cdot (1/3)^{10-2}\) P(2) ≈ 0.00153

Calculate the probability of five heads

Using the binomial probability formula, let x = 5 and n = 10: P(5) = \(\binom{10}{5} \cdot (2/3)^5 \cdot (1/3)^{10-5}\) P(5) ≈ 0.1360

Calculate the probability of eight heads

Using the binomial probability formula, let x = 8 and n = 10: P(8) = \(\binom{10}{8} \cdot (2/3)^8 \cdot (1/3)^{10-8}\) P(8) ≈ 0.3018 The probabilities for each specific outcome are: a. no heads: P(0) ≈ 0.000017 b. two heads: P(2) ≈ 0.00153 c. five heads: P(5) ≈ 0.1360 d. eight heads: P(8) ≈ 0.3018

Key concepts

Weighted Coin Probability

When discussing weighted coin probability, we delve into the fascinating world of biased probabilities where not all outcomes are equally likely. Imagine tossing a coin that’s been tampered with so that the likelihood of landing on heads is significantly higher than that of tails. In the exercise provided, we find a scenario where a coin is weighted such that the probability of heads (P(H)) is two times greater than tails (P(T)). To determine these probabilities, a simple equation based on the concept that the sum of the probabilities for all possible outcomes must equal one is used. After solving, we learn that P(H) is \(\frac{2}{3}\) and P(T) is \(\frac{1}{3}\).

Understanding the weighted coin probability is crucial, especially when it comes to predicting outcomes over multiple coin tosses. As the probabilities differ from the standard 50-50 chances in a fair coin toss, calculating the likelihood of getting a sequence of heads or tails becomes a different ball game—which leads us into the amazing applications of the binomial probability formula.

Probability of Heads and Tails

In any coin toss situation, the probability of heads and tails defines the chance of each side landing face up after a flip. For a fair coin, this is straightforward, at 50% for each side. However, with a weighted coin, as seen in the exercise, these probabilities are skewed. To calculate the probabilities of specific outcomes over several tosses, we must first understand the individual probabilities of heads and tails, which were calculated as \(\frac{2}{3}\) and \(\frac{1}{3}\) respectively in this case.

These individual probabilities become critical components when plugged into the binomial formula to find the probability of a specific number of heads or tails occurring in a sequence of flips. For students or anyone learning probability, grasping this foundational concept is essential, as it sets the stage for more complicated probability distributions and scenarios.

Binomial Theorem Applications

The binomial theorem has intriguing applications in the field of probability—particularly regarding sequences of independent events, like flipping coins. The theorem provides a way to expand expressions that involve powers of sums, which underpins the binomial probability distribution model used in scenarios like our weighted coin problem. Using the binomial probability formula, \( P(x) = \binom{n}{x} \cdot p^x \cdot (1-p)^{n-x} \), where \( n \) is the number of trials, \( x \) is the number of successes, and \( p \) is the probability of success, we can calculate the likelihood of observing a specific number of heads across a series of coin flips.

Through this lens, we can directly calculate the chances of getting no heads, two heads, five heads, or eight heads in ten tosses. By harnessing the binomial theorem, we can universally apply this method to scenarios beyond coin flips, such as predicting genetic traits in biology, financial asset movements in economics, or even the number of correct guesses on a multiple-choice test.

Combinatorics in Probability

Combinatorics is a mathematical domain focused on counting, combination, and permutation of sets—playing an essential role in probability problems, including those involving a binomial distribution. In our coin toss scenario, we use combinatorial calculations to find out how many distinct ways we can achieve a certain number of heads. This is where the \( \binom{n}{x} \) in the binomial formula comes into play, representing the number of combinations of \( n \) items taken \( x \) at a time.

Understanding how to calculate these combinations aids in visualizing how different outcomes can occur. For instance, getting exactly two heads in ten tosses requires recognizing that such two heads can occur in any two flips out of the ten—not just the first two. The combinatory aspect is thus critical in accurately representing the diversity of possible outcomes and is a cornerstone in problems of probability where order does not matter.