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

Let us select five cards at random and without replacement from an ordinary deck of playing cards. (a) Find the pmf of \(X\), the number of hearts in the five cards. (b) Determine \(P(X \leq 1)\).

Short Answer

Expert verified
The pmf of \(X\), the number of hearts, is calculated using the combination formula \(P(X=x) = \frac{C(13, x) * C(39, 5-x)}{C(52, 5)}\), and for \(P(X \leq 1)\), sum the probabilities of 0 and 1 heart.

Step by step solution

01

Basics of Combinations

A combination is the number of ways to select items (without regard to order) from a larger group. In a standard deck of 52 cards, there are 13 hearts. So, when you select 5 cards, the number of hearts \(X\) can range from 0 to 5. Using the combination notation, this can be represented as \(C(n, k) = \frac{n!}{k!(n-k)!}\), where \(n\) is the total number of items, \(k\) is the number of items to select, and \(!\) denotes factorial.
02

Computing PMF

The pmf of \(X\), the number of hearts, is given by the number of ways to get \(X\) hearts and (5 - \(X\)) non-hearts divided by the total number of ways to select 5 cards from the deck. Computationally, this is represented as \(P(X=x) = \frac{C(13, x) * C(39, 5-x)}{C(52, 5)}\), for \(x = 0, 1, 2, 3, 4, 5\). Use this formula to find the pmf for \(X\).
03

Calculating \(P(X \leq 1)\)

The probability \(P(X \leq 1)\) is the sum of the probabilities of 0 and 1 heart. So \(P(X \leq 1) = P(X=0) + P(X=1)\). Use the pmf calculated in Step 2 for these calculations.

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

Each bag in a large box contains 25 tulip bulbs. It is known that \(60 \%\) of the bags contain bulbs for 5 red and 20 yellow tulips while the remaining \(40 \%\) of the bags contain bulbs for 15 red and 10 yellow tulips. A bag is selected at random and a bulb taken at random from this bag is planted. (a) What is the probability that it will be a yellow tulip? (b) Given that it is yellow, what is the conditional probability it comes from a bag that contained 5 red and 20 yellow bulbs?

A mode of a distribution of one random variable \(X\) is a value of \(x\) that maximizes the pdf or pmf. For \(X\) of the continuous type, \(f(x)\) must be continuous. If there is only one such \(x\), it is called the mode of the distribution. Find the mode of each of the following distributions: (a) \(p(x)=\left(\frac{1}{2}\right)^{x}, x=1,2,3, \ldots\), zero elsewhere. (b) \(f(x)=12 x^{2}(1-x), 0

Let \(X\) be a random variable of the continuous type that has pdf \(f(x)\). If \(m\) is the unique median of the distribution of \(X\) and \(b\) is a real constant, show that $$E(|X-b|)=E(|X-m|)+2 \int_{m}^{b}(b-x) f(x) d x$$ provided that the expectations exist. For what value of \(b\) is \(E(|X-b|)\) a minimum?

Consider \(k\) continuous-type distributions with the following characteristics: pdf \(f_{i}(x)\), mean \(\mu_{i}\), and variance \(\sigma_{i}^{2}, i=1,2, \ldots, k .\) If \(c_{i} \geq 0, i=1,2, \ldots, k\), and \(c_{1}+c_{2}+\cdots+c_{k}=1\), show that the mean and the variance of the distribution having pdf \(c_{1} f_{1}(x)+\cdots+c_{k} f_{k}(x)\) are \(\mu=\sum_{i=1}^{k} c_{i} \mu_{i}\) and \(\sigma^{2}=\sum_{i=1}^{k} c_{i}\left[\sigma_{i}^{2}+\left(\mu_{i}-\mu\right)^{2}\right]\) respectively.

From a bowl containing 5 red, 3 white, and 7 blue chips, select 4 at random and without replacement. Compute the conditional probability of 1 red, 0 white, and 3 blue chips, given that there are at least 3 blue chips in this sample of 4 chips.

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