Download PROBABILITY FLOW CHART

PROBABILITY FLOW CHART
EXPERIMENT
S = SAMPLE SPACE
(n(S) =TOTAL OUTCOMES IN
SAMPLE SPACE )
A = EVENT (SUBSET OF S)
(n(A) = TOTAL OUTCOMES IN
EVENT)
SIMPLE EVENTS
(&
COMPLEMENTARY
EVENTS)
MUTUALLY
EXCLUSIVE
NON-MUTUALLY
EXCLUSIVE
COMPLEX
EVENTS
INDEPENDENT
CONDITIONAL
Examples:
Simple Event(& its complement): A box contains 2 black balls and 2 red balls. A ball is
selected at random, its color recorded, and then it is replaced. A second ball is then
selected at random, and its color recorded.
Outcomes can be tabulated as shown below:
R1
R2
B1
B2
B1
B1R1
B1R2
B1B1
B1B2
B2
B2R1
B2R2
B2B1
B2B2
R1
R1R1
R1R2
R1B1
R1B2
R2
R2R1
R2R2
R2B1
R2B2
The probability that both balls are black is 4/16 = ¼. The probability that both balls are
not black = 1 – ¼ = ¾.
Complex Events
Mutually Exclusive: A single card is drawn at random from a standard deck of 52 cards.
Let A = the card is a spade
and B = the card is a heart
A and B are mutually exclusive. P(A) = ¼ and P(B) = ¼.
P(A or B) = P(A) + P(B) = ¼ + ¼ = ½
.
Non-mutually exclusive: Two fair dice are rolled.
Let A = the first die shows a two
and B = the sum is 6 or 7
A and B are non-mutually exclusive. P(A) = 6/36 and P(B) = 11/36.
P(A or B) = P(A) + P(B) – P( A and B ) = 6/36 + 11/36 – 2/36 = 15/36.
Conditional: There are 25 fish in a pond. We know that 14 of these fish are males, 5 of
these males are salmon, and there are 8 salmon in the pond. What is the probability that a
randomly chosen fish is a salmon, given that it is a male?
Solution: The condition ‘given that it is a male’ changes the size of the sample space
under investigation. There are 14 males in all, so the size of the relevant sample space is
14. We also know 5 of these 14 are salmon. Hence, the probability that a randomly
chosen fish is a salmon, given that it is a male is 5/14
Independent: When the outcome of one event has no influence on the outcome of a
second event , the two events are said to be independent.
a) A red die and a green die are rolled.
Event A: red die shows odd
Event B: green die show even.
P(A and B) = P(A) * P(B) = ½ * ½ = ¼
b) Box I has two red balls and one black ball. Box II has one red ball and two black
balls.
Event A: red ball is drawn from Box I
Event B: red ball is drawn from Box II
P(A and B) = P(A) * P(B) = 2/3 * 1/3 = 2/9