Content check — Probability (Grinstead and Snell), Chapter 4: Conditional Probability

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Chapter 4: Conditional Probability

8 problems · 0 figures

Exercise 4.2.1 §4.2 source ↗

Pick a point $x$ at random (with uniform density) in the interval $[0,1]$ ⁠. Find the probability that $x > 1 / 2$ ⁠, given that

(a) $x > 1 / 4$ ⁠.

(b) $x < 3 / 4$ ⁠.

(d) $x^{2} - x + 2 / 9 < 0$ ⁠.

Answer

(a) $2 / 3$

(b) $1 / 3$

(d) $1 / 2$

Exercise 4.2.3 §4.2 source ↗

The Acme Super light bulb is known to have a useful life described by the density function $f (t) = .01 e^{- .01 t},$ where time $t$ is measured in hours.

(a) Find the failure rate of this bulb (see Exercise 2.2.6).

(b) Find the reliability of this bulb after 20 hours.

(d) Find the probability that the bulb burns out in the forty-first hour, given that it lasts 40 hours.

Exercise 2.2.6 in the book ↗

Assume that a new light bulb will burn out after $t$ hours, where $t$ is chosen from $[0, \infty)$ with an exponential density $f (t) = λ e^{- λ t} .$ In this context, $λ$ is often called the failure rate of the bulb.

(a) Assume that $λ = 0.01$ ⁠, and find the probability that the bulb will not burn out before $T$ hours. This probability is often called the reliability of the bulb.

(b) For what $T$ is the reliability of the bulb $= 1 / 2$ ⁠?

Answer

(a) $.01$

(b) $e^{- .01 T}$ where $T$ is the time after 20 hours.

(d) $1 - e^{- .01} \approx .010$

Exercise 4.2.5 §4.2 source ↗

Suppose you choose two numbers $x$ and $y$ ⁠, independently at random from the interval $[0,1]$ ⁠. Given that their sum lies in the interval $[0,1]$ ⁠, find the probability that

(a) $| x - y | < 1$ ⁠.

(b) $x y < 1 / 2$ ⁠.

(d) $x^{2} + y^{2} < 1 / 4$ ⁠.

(e) $x > y$ ⁠.

Answer

(a) $1$

(b) $1$

(d) $π / 8$

(e) $1 / 2$

Exercise 4.2.7 §4.2 source ↗

Let $x$ and $y$ be chosen at random from the interval $[0,1]$ ⁠. Show that the events $x > 1 / 3$ and $y > 2 / 3$ are independent events.

Answer

$P (X > \frac{1}{3}, Y > \frac{2}{3}) = \int_{\frac{1}{3}}^{1} \int_{\frac{2}{3}}^{1} d y d x = \frac{2}{9} .$ But $P (X > \frac{1}{3}) P (Y > \frac{2}{3}) =$ $\frac{2}{3} \cdot \frac{1}{3}$ ⁠, so $X$ and $Y$ are independent.

Exercise 4.2.11 §4.2 source ↗

A coin has an unknown bias $p$ that is assumed to be uniformly distributed between 0 and 1. The coin is tossed $n$ times and heads turns up $j$ times and tails turns up $k$ times. We have seen that the probability that heads turns up next time is $\frac{j + 1}{n + 2} .$ Show that this is the same as the probability that the next ball is black for the Polya urn model of Exercise 4.1.20. Use this result to explain why, in the Polya urn model, the proportion of black balls does not tend to 0 or 1 as one might expect but rather to a uniform distribution on the interval $[0,1]$ ⁠.

Exercise 4.1.20 in the book ↗

The Polya urn model for contagion is as follows: We start with an urn which contains one white ball and one black ball. At each second we choose a ball at random from the urn and replace this ball and add one more of the color chosen. Write a program to simulate this model, and see if you can make any predictions about the proportion of white balls in the urn after a large number of draws. Is there a tendency to have a large fraction of balls of the same color in the long run?

Answer

If you have drawn $n$ times (total number of balls in the urn is now $n + 2$ ⁠) and gotten $j$ black balls, (total number of black balls is now $j + 1$ ⁠), then the probability of getting a black ball next time is $(j + 1) / (n + 2)$ ⁠. Thus at each time the conditional probability for the next outcome is the same in the two models. This means that the models are determined by the same probability distribution, so either model can be used in making predictions. Now in the coin model, it is clear that the proportion of heads will tend to the unknown bias $p$ in the long run. Since the value of $p$ was assumed to be unformly distributed, this limiting value has a random value between 0 and 1. Since this is true in the coin model, it is also true in the Polya Urn model for the proportion of black balls.(See Exercise 20 of Section 4.1.)

Exercise 4.3.1 §4.3 source ↗

One of the first conditional probability paradoxes was provided by Bertrand. (J. Bertrand, Calcul des Probabilités, Gauthier-Uillars, 1888.) It is called the Box Paradox. A cabinet has three drawers. In the first drawer there are two gold balls, in the second drawer there are two silver balls, and in the third drawer there is one silver and one gold ball. A drawer is picked at random and a ball chosen at random from the two balls in the drawer. Given that a gold ball was drawn, what is the probability that the drawer with the two gold balls was chosen?

Answer

$2 / 3$

Exercise 4.3.3 §4.3 source ↗

In the preceding exercise, it is natural to ask “How do we get the information that the given hand has an ace?" Gridgeman considers two different ways that we might get this information. (Again, assume the deck consists of eight cards.)

(a) Assume that the person holding the hand is asked to “Name an ace in your hand" and answers “The ace of hearts." What is the probability that he has a second ace?

(b) Suppose the person holding the hand is asked the more direct question “Do you have the ace of hearts?" and the answer is yes. What is the probability that he has a second ace?

Answer

(a) Consider a tree where the first branching corresponds to the number of aces held by the player, and the second branching corresponds to whether the player answers ’ace of hearts’ or anything else, when asked to name an ace in his hand. Then there are four branches, corresponding to the numbers 1, 2, 3, and 4, and each of these except the first splits into two branches. Thus, there are seven paths in this tree, four of which correspond to the answer ’ace of hearts.’ The conditional probability that he has a second ace, given that he has answered ’ace of hearts,’ is therefore $\frac{[(\frac{48}{12}) + \frac{1}{2} (\frac{3}{1}) (\frac{48}{11}) + \frac{1}{3} (\frac{3}{2}) (\frac{48}{10}) + \frac{1}{4} (\frac{3}{3}) (\frac{48}{9})] / (\frac{52}{13})}{[(\frac{51}{12}) / (\frac{52}{13})]} \approx .6962 .$ (b) This answer is the same as the second answer in Exercise 2, namely $.5612$ ⁠.

Exercise 4.3.5 §4.3 source ↗

Using the notation introduced in Example 4.29, let $p_{x} = {\begin{matrix} \frac{2}{3} (\frac{1}{3})^{k}, & if x = 2^{k}, \\ 0, & otherwise. \end{matrix}$ Show that there is exactly one value of $x$ such that if your envelope contains $x$ ⁠, then you should switch.

Example 4.29 in the book ↗

Suppose that we have two envelopes in front of us, and that one envelope contains twice the amount of money as the other (both amounts are positive integers). We are given one of the envelopes, and asked if we would like to switch.

As above, we let $X$ denote the smaller of the two amounts in the envelopes, and let $p_{x} = P (X = x) .$ We are now in a position where we can calculate the long-term average winnings, if we switch. (This long-term average is an example of a probabilistic concept known as expectation, and will be discussed in Chapter 6.) Given that one of the two sample points has occurred, the probability that it is the point $(x, x / 2)$ is $\frac{p_{x / 2}}{p_{x / 2} + p_{x}},$ and the probability that it is the point $(x, 2 x)$ is $\frac{p_{x}}{p_{x / 2} + p_{x}} .$ Thus, if we switch, our long-term average winnings are $\frac{p_{x / 2}}{p_{x / 2} + p_{x}} \frac{x}{2} + \frac{p_{x}}{p_{x / 2} + p_{x}} 2 x .$ If this is greater than $x$ ⁠, then it pays in the long run for us to switch. Some routine algebra shows that the above expression is greater than $x$ if and only if

$\frac{p_{x / 2}}{p_{x / 2} + p_{x}} < \frac{2}{3} .$

It is interesting to consider whether there is a distribution on the positive integers such that the inequality 4.6 is true for all even values of $x$ ⁠. Brams and Kilgour (ibid.) give the following example.

We define $p_{x}$ as follows: $p_{x} = {\begin{matrix} \frac{1}{3} (\frac{2}{3})^{k - 1}, & if x = 2^{k}, \\ 0, & otherwise. \end{matrix}$ It is easy to calculate (see Exercise 4) that for all relevant values of $x$ ⁠, we have $\frac{p_{x / 2}}{p_{x / 2} + p_{x}} = \frac{3}{5},$ which means that the inequality 4.6 is always true.

Equation 4.6 in the book ↗

$\frac{p_{x / 2}}{p_{x / 2} + p_{x}} < \frac{2}{3} .$

Exercise 4.3.4 in the book ↗

Using the notation introduced in Example 4.29, show that in the example of Brams and Kilgour, if $x$ is a positive power of 2, then $\frac{p_{x / 2}}{p_{x / 2} + p_{x}} = \frac{3}{5} .$

Answer

Let $x = 2^{k}$ ⁠. It is easy to check that if $k \geq 1$ ⁠, then $\frac{p_{x / 2}}{p_{x / 2} + p_{x}} = \frac{3}{4} .$ If $x = 1$ ⁠, then $\frac{p_{x / 2}}{p_{x / 2} + p_{x}} = 0 .$ Thus, you should switch if and only if your envelope contains 1.

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Exercises and answers are adapted from Grinstead and Snell's Introduction to Probability by The CHANCE Project, which is free to read and download, and are used under the GNU FDL licence. They are taken from the GNU version’s LaTeX source (4 July 2006), with answers from the authors’ published answers to the odd-numbered exercises, and set for the web, so they may differ slightly from the printed page. This adaptation is released under the same licence and is neither affiliated with nor endorsed by Grinstead and Snell or American Mathematical Society. Copyright (C) 2006 Peter G. Doyle. This work is a version of Grinstead and Snell's ‘Introduction to Probability, 2nd edition’, published by the American Mathematical Society, Copyright (C) 2003 Charles M. Grinstead and J. Laurie Snell.