Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Galing galing galing galing, Summaries of Law

The best of all the best of all the best of all

Typology: Summaries

2021/2022

Uploaded on 01/26/2023

BeautyBeasty
BeautyBeasty 🇨🇦

3 documents

1 / 4

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
THERESE ANNE Z. CATRAL STAT. ANALYSIS WITH SOFTWARE APP.
BSA - 3 NOVEMBER 26, 2022
QUIZ 2
1. The sample space S is all the ordered pairs of the two whole numbers, the first
from one to three and the second from one to four (Example: 1,4)).
a. S = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)}
Let event A = the even numbers and event B = numbers greater than 13.
b. A = {(1,1), (1,3), (2,2), (2,4), (3,1), (3,3)}
B = {(2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)}
c. P(A) = ,P(B) =
1
22
3
d. A B = {(2,2), (2,4), (3,1), (3,3)}
A B = {(1,1), (1,3),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4)}
e. P(A B) = , P(A B) =
1
35
6
f. B′ = {(1,1),(1,2),(1,3),(1,4)}, P(B′) = 1
3
g. P(A) + P(A′) = 1
h. P(A|B) = =P(B|A) = =
𝑃(𝐴 ∩ 𝐵)
𝑃(𝐵) 1
2𝑃(𝐴 ∩ 𝐵)
𝑃(𝐵) 2
3
Are the probabilities equal? No.
pf3
pf4

Partial preview of the text

Download Galing galing galing galing and more Summaries Law in PDF only on Docsity!

THERESE ANNE Z. CATRAL STAT. ANALYSIS WITH SOFTWARE APP. BSA - 3 NOVEMBER 26, 2022 QUIZ 2

  1. The sample space S is all the ordered pairs of the two whole numbers, the first from one to three and the second from one to four (Example: 1,4)). a. S = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)} Let event A = the even numbers and event B = numbers greater than 13. b. A = {(1,1), (1,3), (2,2), (2,4), (3,1), (3,3)} B = {(2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)} c. P(A) = , P(B) = 1 2 2 3 d. A ∩ B = {(2,2), (2,4), (3,1), (3,3)} A ∪ B = {(1,1), (1,3),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4)} e. P(A ∩ B) = , P(A B) = 1 3

5 6 f. B′ = {(1,1),(1,2),(1,3),(1,4)}, P(B′) = 1 3 g. P(A) + P(A′) = 1

h. P(A|B) = = P(B|A) = =

𝑃(𝐴 ∩ 𝐵) 𝑃(𝐵) 1 2 𝑃(𝐴 ∩ 𝐵) 𝑃(𝐵) 2 3 Are the probabilities equal? No.

  1. You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) of that suit. Three cards are picked at random. a. Suppose you know that the picked cards are Q of spades, K of hearts and Q of spades. Can you decide if the sampling was with or without replacement? ANSWER : With replacement b. Suppose you know that the picked cards are Q of spades, K of hearts, and J of spades. Can you decide if the sampling was with or without replacement? ANSWER : Without replacement
  1. Find P(L ∩ C). ANSWER: P(LC) = P(C|L) * P(L) = 0.550.376 = 0.2068 or 20.68%*
  2. In words, what is L ∩ C? ANSWER: Possibility that the person selected is a Latino registered voter in California who supports life in life in prison without parole of parole over the death penalty for somebody convicted of first-degree murder.
  3. Are L and C independent events? Show why or why not. ANSWER: No, because P(C) is not equal to P(C|L).
  4. Find P(L ∪ C). ANSWER: P(LC): P(C) + P(L) - P(LC) = 0.48 + 0.376 - 0.2068 = 0.6492 or 64.92%
  5. In words, what is L ∪ C? ANSWER: Possibility that the person identified is a Latino Californian or a confirmed California voter who supports life in life in prison without parole of parole over the death penalty for someone found guilty of first-degree murder.
  6. Are L and C mutually exclusive events? Show why or why not. ANSWER: No, because P (LC) is not zero.