study unit 2.1 basic probability

Lesson 2 Probability

Study Unit 2.1Basic ProbabilityFor SQT300SBy B. [email protected] probability formulaProbability definitionsProbability theoremsPermutations and CombinationsBinomial Probability DistributionHypergeometric Probability Distribution2014B. C. Swartz, DISE, CPUT2Basic Probability Formula

2014B. C. Swartz, DISE, CPUT3

Coin, probability for heads is

Die, probability for two is 1/6

Cards, probability for clubs is 13/52

Cards, probability for a seven is 4/52

(Refer to Theorem 1, Theorem 2 and Theorem 5)Some definitionsAn Experiment: A situation involving chance or probability that leads to results (results = outcomes). Outcome also known as an EventEvent: Each possible outcome of a variable is referred to as an event. A simple event is described by a single characteristic. Also said to be one or more outcomes of an experiment.A Joint event: This is an event that has two or more characteristics.The Complement of an event: This includes all events that are not part of that event e.g. If event A is represented by the symbol A, the complement of event A is represented by the symbol A Outcome: Result of a single trial of an experiment2014B. C. Swartz, DISE, CPUT4Theorem 1Probability is ALWAYS expressed as a number between 0.000 and 1.000This value reflects the certainty associated to the likelihood that an event will occurIf probability is 1.000, then there is an absolute certainty that the event will occurIf the probability is 0.000, then there is no certainty that the event will occur2014B. C. Swartz, DISE, CPUT5Theorem 2If P(A) is the probability that event A will occur in an trial, then the probability that A will not occur:P(A) = 1 P(A)

In other words: The probability of the event not occurring = Value of certainty the probability of the event occurring2014B. C. Swartz, DISE, CPUT6More definitionsCollectively exhaustive: An experiment or trial will have set of all outcomes. If these outcomes are a finite set, and if one of the outcomes must occur, then that set is collectively exhaustive. (e.g. Tossing a coin, it will either return a heads or a tail. Thus the possible outcomes are H or T. This set of outcomes are collectively exhaustive)

Mutually exclusive: This term used to describe the relationship between two events in an experiment or trial. Two events (A and B) are mutually exclusive if they cannot occur at the same time. (i.e.) They have no outcomes in common.

2014B. C. Swartz, DISE, CPUT7Theorem 2 cont.Example:If the probability of finding an error on an income tax return is 0.04, what is the probability of finding an error-free or conforming return

P(A) = 1.000 P(A)= 1.000 0.040= 0.960

Therefore, the probability of finding a conforming tax return is 0.960Thus the simple probability of the an event not happening, is 1 the probability that it happens.

In this case, the set of (the tax returns with errors and tax returns without errors) are collectively exhaustiveAlso, the two events, (the tax returns with errors and tax returns without errors), are mutually exclusive2014B. C. Swartz, DISE, CPUT8More examples of Mutually ExclusiveExample of a mutually exclusive event is when a single card is chosen from a deck of 52 playing cards, the probability of choosing a 5 or a King

Another: Rolling a diceProbability that it lands on either an even or an odd number(Refer to Theorem 3 the probability that either of the above-mentioned events will occur is the sum of their respective probabilities)


Theorem 3If A and B are mutually exclusive events (they have no outcomes in common), then the probability of either A or B occurring in the same event, will be the sum of their respective probabilities

P(A or B) = P(A) + P(B)

Whenever the word OR is verbalized, the mathematical operation is addition

2014B. C. Swartz, DISE, CPUT10Reference Table 1

2014B. C. Swartz, DISE, CPUT11Theorem 3 cont.Example:P(X or Z) = P(X) + P(Z) If the 261 parts described in table 1 are contained in a box, what is the probability of selecting a random part produced by supplier X or by supplier Z? P(X or Z) = P(X) + P(Z) = P(53/261) + P(77/261) = 0.498What is the probability of selecting a nonconforming part from supplier x or a conforming part from supplier Z? P(nc. X or co. Z) = P(nc. X) + P(co. Z) = (3/261) + (75/261)= 0.299TAKE NOTE TO READ THE QUESTION CAREFULLY AND DETERMINE WHAT EXACTLY IS BEING ASKED!2014B. C. Swartz, DISE, CPUT12Theorem 4If event A and event B are not mutually exclusive events, (Thus :the events can occur with a common outcome/feature), then the probability that either event A or event B or both occurs is given by

P(A or B or both) = P(A) + P(B) P(both)2014B. C. Swartz, DISE, CPUT13More definitionsNot mutually exclusive: When two events have an outcome in common. That is the possibility exists that the events (A and B) can occur at the same time in the same trial or experiment.

Example of NOT mutually exclusive is when a single card is chosen from a deck of 52 playing cards, that is either a club or a king

Rolling a die, probability of rolling a 5 or an odd number

(Refer to Theorem 4 the probability the either of the above-mentioned events, or both will occur is given by adding the sum of the individual probabilities of the two events and subtracting the common feature shared by the two events)


Theorem 4 cont.Example:

Refer back to Table 1:

What is the probability that a randomly selected part will be from supplier X or a nonconforming unit?

P(X or nc. or both) = P(X) + P(nc.) P(X and nc.)= (53/261) + (11/261) (3/261)= 0.2342014B. C. Swartz, DISE, CPUT15Theorem 5The sum of all the probabilities of a particular event (experiment or trial) is equal to 1.000P(A) + P(B) + . . . +P(N) = 1.000

Understanding this theorem: One needs to know the probability of all other events (A, B...etc) to determine an unknown probability.

Stated differently: Unknown Probability = 1.000 P(the rest)

2014B. C. Swartz, DISE, CPUT16Theorem 5 cont.Example: A health inspector examines 3 products in a subgroup to determine if they are acceptable. From past experience it is known (information given to the inspector), that the probability of finding no nonconforming units in the sample of 3 is 0.990, the probability of finding 1 nonconforming unit in the sample of 3 is 0.006 and the probability of finding 2 nonconforming units in the sample of 3 is 0.003. This probabilities of those events are given = Known Outcomes. He then conducts his own trial and the results of this will be Experimental outcomes. (However the results of this his own trial is still unknown)It is thus known the probability of good products is 99% (or 99 out of 100)

What is the probability of finding 3 nonconforming units in the sample of 3?

There are four events and only four, 0nc.units, 1nc unit, 2nc units and 3nc units P(0) + P(1) + P(2) + P(3) = 1.0000.990 + 0.006 + 0.003 + P(3) = 1.000P(3) = 0.001Thus , the probability of 3 nc. Units in the sample of 3 is 0.0012014B. C. Swartz, DISE, CPUT17Some more on Known Outcomes v.s. Experimental OutcomesSame probability formulaImportant to just be able to recognise the differenceKnown outcomes based on existing dataExperimental outcomes based on trials or experiments being conducted2014B. C. Swartz, DISE, CPUT18Known Outcomes vs Experimental OutcomesA part is selected at random from a container of 50 parts that are known to have 10 nonconforming units. The part is returned to the container, and a record of the number of trials and the number of nonconforming is maintained. After 90 trails, 16 nonconforming units were recorded. What is the probability based on known outcomes and on experimental outcomes?

2014B. C. Swartz, DISE, CPUT19Known outcomes:

Experimental outcomes:

Theorem 6If A and B are independent events, then the probability of both A and B occurring is the product of their respective probabilities

P(A and B) = P(A) x P(B)

Whenever the word AND is verbalized, the mathematical operation is multiplication2014B. C. Swartz, DISE, CPUT20More definitionsIndependent event: Two events, A and B are independent if the event A can occur without affecting the probability of event B occurring. (Either in the same trial or experiment or in successive trials or experiments)

For independent events, only if both events must be considered, then when calculating the probability of both the events occurring, one must take note that this is a form of conditional probability i.e. Event B occurs on condition that Event A has already occurred. Furthermore this above example is also a form of marginal probability (i.e. Joint outcomes)

The test for independence is: P(AB) = P (A). If this is true, then the events are considered to be independent.2014B. C. Swartz, DISE, CPUT21Independent EventExample of an independent event is when tossing a coin AND rolling a 5 on a single 6 sided dieOrRolling a 4 on a single 6 sided die AND then rolling a 1 on a second roll of the same die


Independent EventAnother example: Choosing a 3 from a deck of 52 (Event A), then replacing the card AND then choosing an ace as the second card (Event B).

When two events are independent, (can occur in sequence), one needs to find the probabilities separately, then multiply.

The multiplication is represented by the word AND

(Refer to theorem 6: The probability of two or more independent events occurring is the product of their respective probabilities)2014B. C. Swartz, DISE, CPUT23Theorem 6 cont.Example:

From table 1, what is the probability that 2 randomly selected parts will be from supplier X and supplier Y? Assume that the first part is returned to the box before the second part is selected (called with replacement).

P(X and Y) = P(X) * P(Y)= (53/261) * (131/261)= 0.102

It is low if you consider other possibilities such as XX, YY, ZZ, YX, XZ, ZX, YZ,and ZY. This theorem is also applicable to more than 2 events. 2014B. C. Swartz, DISE, CPUT24Theorem 7If event A and event B are dependent events (thus one event which occurs has an influence on the other event taking place), the probability of both events occurring (A and B) is given as the product of the probability of the event A, and the probability of B under the given condition that A already occurred.P (A and B) = P(A) x P(BIA)2014B. C. Swartz, DISE, CPUT25 And even more definitionsExample of dependent event is when a card is which is a queen is chosen at random from a deck of 52 playing cards (Event A). Without replacing the first card, a second card, which is a jack is chosen (Event B).

Take note: To work out the probability that both of these events took place is a form of Marginal probability (i.e. Set of joint probabilities).

HOWEVER, in order to calculate the probability that just the second event took place one must consider that the first event has an influence on the second event.

Thus, Two events are dependent if the outcome, or occurrence of the first event affects the outcome or occurrence of the second event, (i.e.) After the first event occurred, the probability of the second event changes


Dependent eventsChoosing a jack on the second pick (Event B), given that a queen was chosen on the first pick (Event A) is called conditional probability due to the number of cards are less in the second pick.

The conditional probability of an event B in relationship to event A, is the probability that event B occurs, given that event A has already occurred. The notation for conditional probability is P(BA) means probability of jack being picked, on condition given that another card is being picked out of the deck before event B.

The test for dependence is P(BA) P (B). If this is true, then the two events are not independent.

(Refer to Theorem 7: If first and second events are dependent the probability of first event AND (this means multiply) second event occurring is the product of the two events. However the second event is conditional, thus the probability of that event changes after the first event took place.)2014B. C. Swartz, DISE, CPUT27Theorem 7 cont.Example:From table 1, considering the preceding example used to explain Theorem 6, consider that the first part was not returned to the box before the second part was selected. What is the probability then? P(X and Y) = P(X) * P(YX)= (53/261) * (131/260)= 0.102Since the first part was not returned to the box, there was a total of only 260 parts in the box. What is the probability of choosing both parts from supplier Z?P(Z and Z) = P(Z) * P(ZZ)= (77/261) * (76/260)0.086Since the first part was from supplier Z, there are only 76 from supplier Z of the new total of 260 in the box.The probability of the second event depends on the result of the first event. It is also applicable to more than two events2014B. C. Swartz, DISE, CPUT28Combining theorems......Combination of theorems:

If the 261 parts described in table 1 are contained in a box, what is the probability that two randomly selected parts (with replacement) will have one conforming part from supplier X and one nonconforming part from supplier Y or supplier Z?

P[co.X and (co.Y or co.Z)] = P(co.X) [P(co.Y) + P(co.Z)]= (50/261) * [(6/261) + (2/261)] = 0.192 * [0.023+ 0.008]= 0.005952014B. C. Swartz, DISE, CPUT29Permutations and CombinationsCombination: A loose arrangement of a set of objects.Permutation: An ordered arrangement of a set of objects.With combinations, the order is not important, with permutations, the order is importantA Permutation is an ordered Combination. For a set of objects there will always be more permutations possible than combinations.2014B. C. Swartz, DISE, CPUT30Permutations A permutation is an ordered arrangement of a set of objects. The permutations of the word cup are cup, cpu, upc, ucp, puc and pcu. In this case there are 3 objects in the set and we arrange them into groups of 3 to obtain six permutations.


Permutations cont.How many permutations would there be for 4 objects taken 2 at a time. Use the word fork to represent the four objects, the permutation are fo, of, fr, rf, fk, kf, or, ro, ok, ko, rk, kr.Naturally, if you use the formula and your calculator it is a whole lot easier!


Permutations cont.A witness to a hit and run accident remembered the first 3 digits of the license plate out of 5 digits and noted the fact that the last 2 were numerals. How many owners of automobiles would the police have to investigate?2014B. C. Swartz, DISE, CPUT33AB= (10)(10) = 100If the last 2 were letters, how many would need to be investigatedAB= (26)(26)= 676Suppose the witness remembers further that the numerals were not the same.AB= (10)(9)= 90 or

CombinationsThe order is not important:2014B. C. Swartz, DISE, CPUT34Using the word fork again

Combinations cont.An interior designer has 5 different colour chairs and will use 3 in a living room arrangement. How many diff. Combinations are possible?


Example QuestionsA club consists of 12 members.1.How many slates of officers could be formed if the offices of president, vice president, and secretary are to be filled?2. How many committees of three could be formed to plan the Christmas party?3.If the club consists of nine women and three men, how many ways could the men and women be seated in a row of 12 chairs (for example, M, W, W, W, M, M, W, W,W, W, W, W).2014B. C. Swartz, DISE, CPUT36SolutionsQuestion 1 is a permutation. The order of the people is important

Question 1 is a combination. The order of the people is not important

Question 3 is also a combination. A two letter word problem


Sheet1Inspection results by supplierSupplierNumber conformingNumber NonconformingTotalX50353Y1256131Z75277Total25011261

Sheet2

Sheet3

study unit 2.1 basic probability

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