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

- 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: \(\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}\).

- 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: \(P(x=0) = \frac{1}{9}\), \(P(x=1) = \frac{2}{9} + \frac{2}{9} = \frac{4}{9}\), \(P(x=2) = \frac{4}{9}\).

- 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: \(\bar{x} = 0 \times \frac{1}{9} + 1 \times \frac{4}{9} + 2 \times \frac{4}{9} = 0 + \frac{4}{9} + \frac{8}{9} = \frac{12}{9} = \frac{4}{3}\).

- 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 \(\text\sigma = \text\sqrt{\text\sigma^2} = \text\sqrt{\frac{4}{9}} = \frac{2}{3}\).

- 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.