Lesson 75, 76 and 77

Practice tests and Chapter 7 test

Lesson 74- Using the Binomial Theorem to Calculate Probabilities

Investigate:

P (exactly x correct guesses in 6 rolls) = 6Cx (1/6)^x (5/6) ^6-x
note: The # of pathways with x C's times the probability of a specific pathway with x C's
Example 1:
P(exactly 6 correct guesses in 6 rolls) = 6C6(1/6)^6= 0.0000214
Note: calculations like these that use the word exactly you may use the function call Binompdf, put in the # of plays, chances, guesses
Binompdf (6,1/6,6)= 0.0000214

If the question says at least or at most in the question you must use Binomcdf:
P(at most to heads in 12 flips)= binomcdf( 12, 0.5, 2)= 0.0192

Binomial Experiment:
A binomial experiment consists of:
- a fixed number of independent and identical trials
- only two possible outcomes ( success or failure) per trail

Binomial Probability:

The probability of getting x successes and n-x  failures in n trials is
P(x)= nCx p^x q^n-x          P=success   Q= failure

Lesson 73- Using Permutations and Combinations to Calculate Probabilities

Using Combination's:
Sample space is list of all possible ways (equally likely) to select 2 cards from 52 cards:

Number of unordered ways to select 2 cards from 52 cards:  52C2 = 1326
Number of unordered ways to select 2 aces:  4C2 = 6

P(both cards aces) =  4C2    =  1/ 221
                                 52C2

Using Permutations:
Number of ordered ways to select 2 cards from 52 cards: 52P2

Number of ways to select 2 aces: 4P2

P(both cards aces)= 4P2   = 1/221
                              52P2

Example 1:    Three prizes are awarded in a raffle. One hundred people hold a ticket each

a) what is the probability that Alice, Ben, and Concetta win first, second, and third prizes respectively?
* Order Does Matter*
   1      = 1.03 x 10^-6
100P3

b) What is the probability that Alice, Ben and Concetta are the 3 prize winners (though not necessarilyin that order)?
* Order Does Not Matter*
      1   =  6.18 x 10 ^ -6
100C3

* Order Does Matter* = Permutation       * Order Does Not Matter*= Combination

Lesson 72- Problems Involving Conditional Probability

Conditional Probability:
Probability of given  A in the event of B  
Formula: P(a | b)

Example 1:
A card is drawn from a shuffled deck for 52 cards, and not replaced. Then a second card is drawn.
a)  What is the probability  that the second card drawn is a king?
P( k and k ) +  P( K(with line on top) and K)
4/52 x 3/51 +  48/52 x 4/51 = 1/13
* use a tree diagram

Lesson 71- The event a and b

When do you multiply probabilities?
Situation 1: Drawing two balls from the pot, without replacement.
A: the first ball is white                                Find: P(a) = 2/3      P(B given that the first ball drawn is white)
B: the second ball is black                                                         = P( b | a) = 1/2

* note: using a tree diagram for this question will be useful!

Multiplication Law:
General Case: P( a and b) = P(a) x P(B|A)

Equivalently:  p(b | a) = P (A and B)
                                       P(A)
Special Sub-case:  Events A and B are independent if p(B | A) = P(b)
                            For independent events: P(a and b) = p(a) x p(b)
Two events that are not independent are called dependent

Example:
Three cards are drawn with replacement from a shuffled deck of 52 cards
What is the probability that all three are spades?
P (S and S and S) = 13/52 x 12/51 x 11/50 = 11/850

Lesson 70- The event A or B

Addition Law:
a) General  Case:
               P (A or B) = P(A) + P(B) - P(A and B)                * Overlap
b) Special Sub-case:  
 If A and B are mutually exclusive then
    P(A and B) = P(A) + P(B)

Example 1:
One card is randomly drawn from deck of 52 cards
Define the following events:

S:  The card is a spade
R:  The card is red
F:  The card is face card

Which of these three are mutually exclusive?
S, R (do not overlap)

Determine the following probabilities directly by counting outcomes:
P(s) = 13/52   P(r)= 26/52    P(s and r)= O    P(s or r)= 13/52 + 26/52 = 39/52
P(s) = 13/52   P(f) = 12/52   P(s and f) = 3/52   P(s or f) = 13/52 + 12/52 - 3/52 = 22/52

*note: using a Venn-diagram with these problems will make things a lot easier, as the info. is right in front of you. 

Lesson 67 - Related Events

A list of all possible (non-overlapping) outcomes of an experiment is called a sample space

Example 1:
One bill is randomly selected from the pot
a) Sample space:
   $0, $5, $10
b) What is the probability that the $5 bill is selected?
P(5)= 1/3

Related Events:
Complement- Opposite
The event that A does not occur is the complement of event A, and is denoted  A(with line on top)
*note: P(A) + P(A)*with line on top = 1

A and B
The event that both A and B occur is denoted by A and B

A or B
The event that A or B occurs (or both) is denoted by A or B

Example 2:
If each of the 13 outcomes in the sample space is equally likely, find:
Sample Space:

P(A)= 4/13 = 30%                              P(B)= 5/13
P(not A)= 9/13= 70%                         P(notB)= 8/13
P(A and B)= 1/13                               P(A or B)= 4/13 + 5/13 - 1/13 = 8 /13