Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

1. Consider a daily lottery as described in Example 5.5.4.

a. Compute the probability that two particular days ina row will both have triples.

b. Suppose that we observe triples on a particular day. Compute the conditional probability that we observe triples again the next day.

Short Answer

Expert verified
  1. 0.0001
  2. 0.01

Step by step solution

01

Given information

A common daily lottery game involves the drawing of three digits from 0 to 9 independently with replacement and independently from day to day. Lottery watchers often get excited when all three digits are the same, an event called triples.

If p is the probability of obtaining triples, and if X is the number of days without triples before the first triple is observed, then X has the geometric distribution with parameter p. In this case, it is easy to see that p = 0.01, since there are 10 different triples among the 1000 equally likely daily numbers.

The number X of daily draws without a tripleuntil we see a triple has the geometric distribution with parameter p = 0.01. \(X \sim Geo\left( {p = 0.01} \right)\)

02

Calculating the probability

a.

The probability that two particular days in a row will both have triples is the same as having two consecutive successes.

\(\begin{array}{c}p\left( x \right) = {\left( {0.1} \right)^2}\\ = 0.0001\end{array}\)

Thus, the required probability is 0.0001.

03

Calculating the conditional probability

b.

The conditional probability is defined as follows:

\(P\left( {A|B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}\)

It is read as Probability A, given B is probability of A and B upon probability of occurrence of B.

In this case,

A=occurrence of a triple the next day

B=occurrence of a triple today

Therefore,

\(\begin{array}{c}P\left( {triple\;tomorrow|triple\;today} \right) = \frac{{P\left( {triple\;today\;and\;tomorrow} \right)}}{{P\left( {triple\;today} \right)}}\\ = \frac{{0.01 \times 0.01}}{{0.01}}\\ = 0.01\end{array}\)

Thus, the conditional probability that we observe triples again the next day given they occurred today is 0.01.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \({{\bf{X}}_{{\bf{1,}}}}{\bf{ }}{\bf{. }}{\bf{. }}{\bf{. , }}{{\bf{X}}_{\bf{n}}}\)be i.i.d. random variables having thenormal distribution with mean \({\bf{\mu }}\) and variance\({{\bf{\sigma }}^{\bf{2}}}\). Define\(\overline {{{\bf{X}}_{\bf{n}}}} {\bf{ = }}\frac{{\bf{1}}}{{\bf{n}}}\sum\limits_{{\bf{i = 1}}}^{\bf{n}} {{{\bf{X}}_{\bf{i}}}} \) , the sample mean. In this problem, weshall find the conditional distribution of each \({{\bf{X}}_{\bf{i}}}\)given\(\overline {{{\bf{X}}_{\bf{n}}}} \).

a.Show that \({{\bf{X}}_{\bf{i}}}\)and\(\overline {{{\bf{X}}_{\bf{n}}}} \) have the bivariate normal distributionwith both means \({\bf{\mu }}\), variances\({{\bf{\sigma }}^{\bf{2}}}{\rm{ }}{\bf{and}}\,\,\frac{{{{\bf{\sigma }}^{\bf{2}}}}}{{\bf{n}}}\),and correlation\(\frac{{\bf{1}}}{{\sqrt {\bf{n}} }}\).

Hint: Let\({\bf{Y = }}\sum\limits_{{\bf{j}} \ne {\bf{i}}} {{{\bf{X}}_{\bf{j}}}} \).

Nowshow that Y and \({{\bf{X}}_{\bf{i}}}\) are independent normals and \({{\bf{X}}_{\bf{n}}}\)and \({{\bf{X}}_{\bf{i}}}\) are linear combinations of Y and \({{\bf{X}}_{\bf{i}}}\) .

b.Show that the conditional distribution of \({{\bf{X}}_{\bf{i}}}\) given\(\overline {{{\bf{X}}_{\bf{n}}}} {\bf{ = }}\overline {{{\bf{x}}_{\bf{n}}}} \)\(\) is normal with mean \(\overline {{{\bf{x}}_{\bf{n}}}} \) and variance \({{\bf{\sigma }}^{\bf{2}}}\left( {{\bf{1 - }}\frac{{\bf{1}}}{{\bf{n}}}} \right)\).

Suppose that seven balls are selected at random withoutreplacement from a box containing five red balls and ten blue balls. If \(\overline X \) denotes the proportion of red balls in the sample, what are the mean and the variance of \(\overline X \) ?

Suppose that X has the normal distribution with mean\({\bf{\mu }}\)and variance\({{\bf{\sigma }}^{\bf{2}}}\). Express\({\bf{E}}\left( {{{\bf{X}}^{\bf{3}}}} \right)\)in terms of\({\bf{\mu }}\)and\({{\bf{\sigma }}^{\bf{2}}}\).

Suppose that the random variables X1 and X2 havea bivariate normal distribution, for which the joint p.d.f.is specified by Eq. (5.10.2). Determine the value of the constant b for which \({\bf{Var}}\left( {{{\bf{X}}_{\bf{1}}}{\bf{ + b}}{{\bf{X}}_{\bf{2}}}} \right)\)will be a minimum.

Prove Theorem 5.5.5.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free