Probability & Discrete Random Variables

Question 1

P(A) = (complex)

Answer 1

• P(A) = P(A ∩ B) + P(A ∩ Bc)

- Probability of A = probability of A & B and probability of A and not B

Question 2

P(A ∪ B) =

Answer 2

• P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

- Intersection must be subtracted because it is counted twice (think Venn)

Question 3

P(A|B) =

Answer 3

P(A|B) = ( P(A ∩ B) ) / P(B)

Question 4

When are two events independent?

Answer 4

Two events are independent if P(A|B) = P(A) or P(B|A) = P(B)

Question 5

If A and B are independent then….

Answer 5

• P(A ∩ B) = P(A)*P(B)
• Then P(A|B)
  
  • = [P(A ∩ B)] / P(B)
  • = [P(A)*P(B)] / P(B)
  • = P(A)

Question 6

What are joint probabilities?

Answer 6

• The joint likelihood of fulfilling multiple criteria
• i.e intersections
• Marginal Probabilities

Question 7

What are marginal probabilities?

Answer 7

• The probability of a single event occuring given information about joint probabilities
• Computed by adding A1 A2 or B1 B2 etc. rule 1
• P(A1) = P(A1 ∩ B1) + P(A1 ∩ B2)
• P(B2) = P(B2 ∩ A1) + P(B2 ∩ A2)

Question 8

What is the multiplication rule used for?

Answer 8

If A and B are not independent, use this rule to find P(A ∩ B)

Question 9

What is the multiplication rule?

Answer 9

• P(A ∩ B) = P(A|B) * P(B)

- and P(A ∩ B) = P(B|A) * P(A)

Question 10

What is the multiplication rule for independent events?

Answer 10

• If A and B are independent, the probability of A and B is equal to the probability of A times the probability of B
• P(A ∩ B) = P(A)P(B)

Question 11

How is marginal probability calculated using the multiplication rule?

Answer 11

• P(A ∩ B) = P(A|B1) * P(B1) + P(A|B2) * P(B2) + … + P(A|Bk) * P(Bk)
• Where B1,B2,…,Bn are k mutually exclusive and collectively exhaustive events

Question 12

How is independence determined for conditional probabilities?

Answer 12

If independent, P(A given B) = P(A)

Question 13

What is the 1st counting rule?

Answer 13

• Determines the number of possible outcomes for a set of mutually exclusive and collectively exhaustive events
• If any one of k different mutually exclusive and collectively exhaustive events can occur on each of n trials, the number of possible outcomes is equal to k^n

Question 14

What is the 2nd counting rule?

Answer 14

• More general version of the first and allows the number of possible events to differ from trial to trial
• If there are k1 events on the first trial, k2 events on the second trial, … , and kn events on the nth trial, then the number of possible outcomes is (k1)(k2)…(kn)

Question 15

What is the 3rd counting rule?

Answer 15

• Involves computing the number of ways that a set of items can be arranged in order
• n! = (n)(n - 1) … (1)
• 0! is defined as 1

Question 16

What is the 4th counting rule?

Answer 16

• In many instances you need to know the number of ways in which a subset of an entire group of items can be arranged in order
• Each possible arrangement is called a permutation
• The number of ways of arranging x objects selected from n objects is nPx = (n!) / (n - x)!

Question 17

What is the 5th counting rule?

Answer 17

• In many situation, you are not interested in the order of the outcomes but only in the number of ways that x items can be selected from n items, irrespective of order
• Each possible selection is called a combination
• The number of ways of selecting x objects from n objects,
  irrespective of order, is equal to nCx = (n!) / x!(n - x)!