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Chapter 15: Problem 8

(a) A weighted coin has probability \(\frac{2}{3}\) of coming up heads and probability \(\frac{1}{3}\) of coming up tails. The coin is tossed twice. Let \(x=\) number of heads. Set up the sample space for \(x\) and the associated probabilities. (b) Find \(\bar{x}\) and \(\sigma\) (c) If in (a) you know that there was at least one tail, what is the probability that both were tails?

Short Answer

Expert verified
Sample space: HH, HT, TH, TT with respective probabilities. \(\bar{x} = \frac{4}{3}\) and \(\text\sigma = \frac{2}{3}\). If at least one tail, probability of both being tails: 0.2.

Step by step solution

01

- Sample Space and Probabilities

The coin is tossed twice, and each toss can result in either heads (H) or tails (T). List all possible outcomes: HH, HT, TH, TT. Calculate the probability for each outcome using the given probabilities for heads and tails. Probability of HH: \(\frac{2}{3} \times \frac{2}{3} = \frac{4}{9}\). Probability of HT: \(\frac{2}{3} \times \frac{1}{3} = \frac{2}{9}\). Probability of TH: \(\frac{1}{3} \times \frac{2}{3} = \frac{2}{9}\). Probability of TT: \(\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}\).
02

- Random Variable x and Probabilities

Define the random variable \(x =\) number of heads. List the possible values of \(x\) and their corresponding probabilities. \(x=0\) for TT, \(x=1\) for HT and TH, and \(x=2\) for HH. Calculate the probabilities: \(P(x=0) = \frac{1}{9}\), \(P(x=1) = \frac{2}{9} + \frac{2}{9} = \frac{4}{9}\), \(P(x=2) = \frac{4}{9}\).
03

- Mean of x (\(\bar{x}\))

Use the probability distribution to calculate the mean \(\bar{x}\). \(\bar{x} = \text{E}(x) = \text{\textbackslash sum } x_i P(x_i)\). Compute: \(\bar{x} = 0 \times \frac{1}{9} + 1 \times \frac{4}{9} + 2 \times \frac{4}{9} = 0 + \frac{4}{9} + \frac{8}{9} = \frac{12}{9} = \frac{4}{3}\).
04

- Standard Deviation of x (\(\boldsymbol{\text\sigma}\))

Calculate the variance \(\text\sigma^2\). \(\text\sigma^2 = \text{E}(x^2) - \bar{x}^2\). Compute \(\text{E}(x^2)\) first: \(\text{E}(x^2) = 0^2 \times \frac{1}{9} + 1^2 \times \frac{4}{9} + 2^2 \times \frac{4}{9} = 0 + \frac{4}{9} + \frac{16}{9} = \frac{20}{9}\). Then calculate \(\text\sigma^2 = \frac{20}{9} - \frac{16}{9} = \frac{4}{9}\). Finally, find the standard deviation \(\text\sigma = \text\sqrt{\text\sigma^2} = \text\sqrt{\frac{4}{9}} = \frac{2}{3}\).
05

- Conditional Probability

Given at least one tail (i.e., HT, TH, TT), find the probability that both are tails (TT). Calculate the probability of getting at least one tail: \(P(\text{at least one T}) = 1 - P(\text{HH}) = 1 - \frac{4}{9} = \frac{5}{9}\). Then, calculate the conditional probability: \(P(\text{TT} | \text{at least one T}) = \frac{P(\text{TT})}{P(\text{at least one T})} = \frac{\frac{1}{9}}{\frac{5}{9}} = \frac{1}{5} = 0.2\).

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Key Concepts

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

Sample Space and Probabilities
In probability, a sample space represents all possible outcomes of an experiment. When we toss a weighted coin twice, each toss could be heads (H) or tails (T). With two coins, the sample space includes all combinations: HH, HT, TH, and TT.
Given the coin's bias, the probability of heads is \(\frac{2}{3}\) and tails is \(\frac{1}{3}\). To calculate the probability of each possible outcome, multiply the probabilities:
  • HH: \(\frac{2}{3} \times \frac{2}{3} = \frac{4}{9}\)
  • HT: \(\frac{2}{3} \times \frac{1}{3} = \frac{2}{9}\)
  • TH: \(\frac{1}{3} \times \frac{2}{3} = \frac{2}{9}\)
  • TT: \(\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}\)
These probabilities tell us the likelihood of each outcome when tossing the coin twice.
Random Variable
A random variable quantifies outcomes of a probabilistic experiment. In our example, define \(x\) as the number of heads in two coin tosses. Possible values for \(x\) are:
  • 0 heads (TT)
  • 1 head (HT, TH)
  • 2 heads (HH)
To find the probability of each \(x\) value:
  • \(P(x = 0) = \frac{1}{9}\)
  • \(P(x = 1) = \frac{2}{9} + \frac{2}{9} = \frac{4}{9}\)
  • \(P(x = 2) = \frac{4}{9}\)
This probability distribution helps us understand the expected number of heads in two tosses.
Mean (Expected Value)
The mean, or expected value \(\bar{x}\), shows the average outcome. Calculate the mean by summing the products of each value and its probability: \(\bar{x} = \text{E}(x) = \text\backslashsum x_i P(x_i)\)
For our example:
\(\bar{x} = 0 \times \frac{1}{9} + 1 \times \frac{4}{9} + 2 \times \frac{4}{9} = 0 + \frac{4}{9} + \frac{8}{9} = \frac{12}{9} = \frac{4}{3}\)
The mean number of heads in two tosses of this weighted coin is \(\frac{4}{3}\).
Standard Deviation
Standard deviation measures the spread of a set of values. First, find the variance \(\text{\textbackslashsigma}^2\) and then take its square root. To calculate variance:
1. Compute \(\text{E}(x^2)\)
2. Find \(\text{\textbackslashsigma}^2 = \text{E}(x^2) - \bar{x}^2\)
For our example:
  • \(\text{\textbackslash{E}}(x^2) = 0^2 \times \frac{1}{9} + 1^2 \times \frac{4}{9} + 2^2 \times \frac{4}{9} = 0 + \frac{4}{9} + \frac{16}{9} = \frac{20}{9}\)
  • \(\text{\textbackslashsigma}^2 = \frac{20}{9} - \frac{16}{9} = \frac{4}{9}\)
Then compute standard deviation:
\(\text{\textbackslashsigma} = \text{\textbackslashsqrt{\frac{4}{9}}} = \frac{2}{3}\)
It tells us how much the number of heads varies around the mean.
Conditional Probability
Conditional probability is the chance of an event occurring given that another event has already happened. In our case, we know there's at least one tail in two tosses. We need the probability that both tosses are tails.
First, find the probability of at least one tail:
\(P(\text{at least one T}) = 1 - P(\text{HH}) = 1 - \frac{4}{9} = \frac{5}{9}\)
Next, calculate the conditional probability that both are tails:
\(P(\text{TT} | \text{at least one T}) = \frac{P(\text{TT})}{P(\text{at least one T})} = \frac{\frac{1}{9}}{\frac{5}{9}} = \frac{1}{5} = 0.2\)
This means there's a 20% chance of both tosses being tails if we know at least one was a tail.

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Most popular questions from this chapter

For the given values of \(n\) and \(p\) computer plot graphs of the binomial density function for the probability of \(x\) successes in \(n\) Bernoulli trials with probability \(p\) of success. $$n=50, p=1 / 5$$

What is the probability that the 2 and 3 of clubs are next to each other in a shuffled deck? Hint: Imagine the two cards accidentally stuck together and shuffled as one card.

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(a) A candy vending machine is out of order. The probability that you get a candy bar (with or without return of your money) is \(\frac{1}{2},\) the probability that you get your money back (with or without candy) is \(\frac{1}{3}\), and the probability that you get both the candy and your money back is \(\frac{1}{12}\). What is the probability that you get nothing at all? Suggestion: Sketch a geometric diagram similar to Figure 3.1, indicate regions representing the various possibilities and their probabilities; then set up a four-point sample space and the associated probabilities of the points. (b) Suppose you try again to get a candy bar as in part (a). Set up the 16 -point sample space corresponding to the possible results of your two attempts to buy a candy bar, and find the probability that you get two candy bars (and no money back); that you get no candy and lose your money both times; that you just get your money back both times.
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