/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Let us select five cards at rand... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 5

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 is calculated using the formula \[p(x) = \frac{C(13, x) \cdot C(39, 5-x)}{C(52, 5)}\] where \(x\) ranges from 0 to 5. The probability of \(P(X \leq 1)\) is calculated by adding the individual probabilities when \(X = 0\) and \(X = 1\).

Step by step solution

01

Bucketing of the Cards

The ordinary deck has 52 cards which includes 4 suits (hearts, diamonds, clubs, spades) each with 13 cards. So, there are 13 hearts and 39 cards which are not hearts.
02

Calculate PMF

To find the PMF of \(X\), the number of hearts in the drawn cards, one needs to calculate the probability for all \(X = 0, 1, 2, 3, 4, 5\). In this case, one can use the concept of combination to find the number of ways to select \(X\) hearts from 13 hearts (\(C(13, X)\)) and (5 - \(X\)) non-hearts from 39 non-hearts (\(C(39, 5-X)\)). Then, one can find the PMF by dividing the product of the number of ways to select hearts and non-hearts by the total number of ways to select 5 cards from the deck (\(C(52, 5)\)).
03

Calculate \(P(X \leq 1)\)

This requires the sum of probabilities when \(X = 0\) and \(X = 1\). Applying the same combination method as previously, the probability when \(X = 0\) (no hearts) can be found through choosing 5 non-hearts from 39, divided by the total number of ways to select 5 cards. Meanwhile, when \(X = 1\), it implies 1 heart out of 13 and 4 non-hearts from 39. Their sum will give \(P(X \leq 1)\).

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 91影视!

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

A French nobleman, Chevalier de M茅r茅, had asked a famous mathematician, Pascal, to explain why the following two probabilities were different (the difference had been noted from playing the game many times): (1) at least one six in four independent casts of a six-sided die; (2) at least a pair of sixes in 24 independent casts of a pair of dice. From proportions it seemed to de M茅r茅 that the probabilities should be the same. Compute the probabilities of (1) and (2).

Let \(C\) denote the set \(\left\\{(x, y, z): x^{2}+y^{2}+z^{2} \leq 1\right\\} .\) Using spherical coordinates, evaluate $$ Q(C)=\iiint_{C} \sqrt{x^{2}+y^{2}+z^{2}} d x d y d z $$

Let a random variable \(X\) of the continuous type have a pdf \(f(x)\) whose graph is symmetric with respect to \(x=c .\) If the mean value of \(X\) exists, show that \(E(X)=c\) Hint: Show that \(E(X-c)\) equals zero by writing \(E(X-c)\) as the sum of two integrals: one from \(-\infty\) to \(c\) and the other from \(c\) to \(\infty\). In the first, let \(y=c-x\); and, in the second, \(z=x-c\). Finally, use the symmetry condition \(f(c-y)=f(c+y)\) in the first.

A secretary types three letters and the three corresponding envelopes. In a hurry, he places at random one letter in each envelope. What is the probability that at least one letter is in the correct envelope? Hint: Let \(C_{i}\) be the event that the ith letter is in the correct envelope. Expand \(P\left(C_{1} \cup C_{2} \cup C_{3}\right)\) to determine the probability.

A median of a distribution of one random variable \(X\) of the discrete or continuous type is a value of \(x\) such that \(P(X
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
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.