Independent Events 48. Medical Experiment. A medical experiment - - PowerPoint PPT Presentation

Independent Events

• 48. Medical Experiment. A medical experiment showed that the probability

that a new medicine is effective is 0.75, the probability that a patient will have a certain side effect is 0.4, and the probability that both events occur is 0.3. Decide whether these events are dependent or independent.

Dan Barbasch Math 1105 Chapter 7.5, 7.6 Week of September 4 1 / 11

Drawing Cards without Replacement I

Example

Draw two cards from a well shuffled deck without replacement. Are the two events A = { first card is an ace } B = { second card is a heart } independent?

Dan Barbasch Math 1105 Chapter 7.5, 7.6 Week of September 4 2 / 11

Drawing Cards without Replacement II

Answer.

For a tree diagram check your class notes. A well shuffled deck means/implies uniform distribution, any pair of two cards is equally likely to be drawn. There are 52 · 51 equally likely choices. For the first card there are four choices of ace: spade, club, heart and

• diamond. For each choice of a first card there are 51 choices of a second
• card. So for each first card, there are 51 choices for the second card. Then

p(A) = 4

52 · 51 51 = 4 52.

For B, there are 13 choices of H, and 39NH choices for the first draw. For each of the 13H choices, there are 12/51 choices of H for the second card. For the 39NH choices, there are 13H choices for the second draw. So p(B) = 13

52 · 12 51 + 39 52 · 13

• 51. Note that p(B) = 13

52 as well. Why?!

For A ∩ B there are four choices of aces for the first card. For the choice

• f ace of hearts there are 12 choices of hearts for the second card. For the
• ther three aces there are 13 choices of hearts for the second card. So

p(A ∩ B) = 1

52 · 12 51 + 3 52 · 13 51.

You can check that p(A ∩ B) = p(A) · p(B).

Dan Barbasch Math 1105 Chapter 7.5, 7.6 Week of September 4 3 / 11

Tree Diagrams I

• 1♦

51 other 1♥ 51 other 1♠ 51 other 1♣ 51 other 48 not ace 51 other P( first draw is an ace ) = 4 × 1 52 · 51 51 One of four aces, each with probability

1

• 52. For the second draw, there are

51 choices for each first draw.

Dan Barbasch Math 1105 Chapter 7.5, 7.6 Week of September 4 4 / 11

• 13H

12H other than first H 39NH 13H P( second draw is a heart ) = 13 × 1 52 · 12 51 + 39 × 1 52 · 13 51 One of 13 hearts, each with probability

1

• 52. For the second draw, there are

12H choices left. In addition, 39NH for the first choice, and 13H out of 51 for the second choice.

Dan Barbasch Math 1105 Chapter 7.5, 7.6 Week of September 4 5 / 11

1♥ 12H 39NH 1♦ 13H 38NH 1♠ 13H 38NH 1♣ 13H 38NH 48NA 51O P( first draw is an ace and second draw is a heart ) = = 1 × 1 52 · 12 × 1 51 + 3 × 1 52 · 13 × 1 51 First Ace is hearts with probability

1 52 times 12 choices of hearts with

probability

1 51

plus 3 choices of Ace each with probability

1 52 times 13 choices of heart each

with probability

1 51; for the second draw, there are 51 choices.

Dan Barbasch Math 1105 Chapter 7.5, 7.6 Week of September 4 6 / 11

Bayes’s Theorem I

Basic Question Given p(E | F), what is p(F | E)? Suppose the probability that a student taking M1105 is known given that the student is in A&S. What is the probability that a student is n A&S given that he/she is taking M1105? Algebra p(E | F) = p(E ∩ F) p(F) p(F | E) = p(E ∩ F) p(E) = p(E ∩ F) p(F) · p(F) p(E) = p(E | F) · p(F) p(E) p(E) = p(E ∩ F) + p(E ∩ F c) = p(E | F) · p(F) + p(E | F c) · p(F c).

Dan Barbasch Math 1105 Chapter 7.5, 7.6 Week of September 4 7 / 11

Example (24 in Section 7.6)

Hepatitis Blood Test. The probability that a person with certain symptoms has hepatitis is 0.8. The blood test used to confirm this diagnosis gives positive results for 90% of people with the disease and 5% of those without the disease. What is the probability that an individual who has the symptoms and who reacts positively to the test actually has hepatitis?

Example

A disease test is advertised as being 99% accurate: if you have the disease, you will test positive 99% of the time, and if you don’t have the disease, you will test negative 99% of the time. If 1% of all people have this disease and you test positive, what is the probability that you actually have the disease?

Answer.

0.01, 0.25, 0.50, 0.75, 0.99?!

Dan Barbasch Math 1105 Chapter 7.5, 7.6 Week of September 4 8 / 11

Counting

Multiplication Principle: Two finite sets A and B. Count the number of pairs {(a, b) : a ∈ A, b ∈ B}. Answer: n(A) × n(B). Permutations: The number of ways you can order n distinct objects. Answer: n! = n · (n − 1) · · · · · 2 · 1. By definition 0! = 1. The number of ways you can select k elements out of n with

• rder

Answer: P(n, k) = n · (n − 1) · · · · · (n − k + 1) =

n! (n−k)!.

Combinations: Number of ways to select k elements out of n without

• rder: .

C(n, k) = n

k

• =

n! (n−k)!k!.

Dan Barbasch Math 1105 Chapter 7.5, 7.6 Week of September 4 9 / 11

Examples I

Example (1)

Four men and three women are to be seated in seven chairs lined up in a row.

1 In how many ways can this be done, assuming the men and women

are individually distinguishable? For example the men are John. David and Bill, and the women are Mary, Ellen and Ann.

2 In how many ways can this be done, assuming the men and women

are not distinguishable? In other words we can only distinguish men from women but not exactly who they are.

Dan Barbasch Math 1105 Chapter 7.5, 7.6 Week of September 4 10 / 11

Examples II

Example (Radio Station Call Letters)

How many different 4-letter radio station call letters can be made if

• a. the first letter must be K or W and no letter may be repeated?
• b. repeats are allowed, but the first letter is K or W?
• c. the first letter is K or W, there are no repeats, and the last letter is R?

Example (YOUR TURN 7)

A student has 4 pairs of identical blue socks, 5 pairs of identical brown socks, 3 pairs of identical black socks, and 2 pairs of identical white socks. In how many ways can he select socks to wear for the next two weeks?

Dan Barbasch Math 1105 Chapter 7.5, 7.6 Week of September 4 11 / 11