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Probabilities Involving Combinations

Example1:  In the Illinois Lottery, an urn contains balls numbered 1-54.  From the urn, 6 balls are randomly chosen without replacement.  For a $1 bet, a player chooses two sets of 6 numbers.  In order to win, all six numbers must match those chosen from the urn.  The order in which the balls are selected does not matter.  What is the probability of winning Lottto?

The probability of winning is given by the number of ways a ticket could win, divided by the size of the sample space.  Each ticket has two sets of six numbers, so there are two chances (for the two sets of numbers) of winning for each ticket.  The size of the sample space S is the number of ways that 6 objects can be selected from 54 objects without replacement and without regard to order, so that  .

 The size of the sample space is

Each ticket has two sets of 6 numbers, so a player has two chances of winning for each $1.  If E is the event “winning ticket”, the   The probability of E is

There is about a 1 in 13,000,000 chance of winning the Illinois Lottery!

To calculate the probability of winning the Illinois Lottery, you must calculate .

To calculate the probability of winning the Illinois Lottery, you must calculate .

Enter the numerator, 2, into your calculator. Next press: and enter the first value in the denominator, 54 press MATH, highlight PRB and select 3:nCr, enter the next value, 6. Press ENTER and the answer will be displayed on your screen.

Notice, that the answer appears in scientific notation. To convert to standard notation, move the decimal point 8 places to the left. The answer is .0000000774.

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Example 2:  A shipment of 120 fasteners that contains 4 defective fasteners was sent to a manufacturing plant.  The quality-control manager at the manufacturing plant randomly selects five fasteners and inspects them.  What is the probability that exactly one of the fasteners is defective?

The probability that exactly one fastener is defective is found by calculating the number of ways of selecting exactly 1defective fastener in 5 fasteners and then dividing this result by the number of ways of selecting 5 fasteners from 120 fasteners.  To choose exactly 1 defective in the 5 requires choosing one defective from the four defectives and 4 nondefectives from the 116 nondefectives.  The order in which the fasteners are selected does not matter, so we use combinations.

The number of ways of choosing 1 defective fastener from 4 defective fasteners is .  The number of ways of choosing four nondefective fasteners from 116 nondefectives is .  Using the Multiplication Principle, we find that the number of ways of choosing 1 defective and 4 nondefective fasteners is

The number of ways of selecting 5 fasteners from 120 fasteners is .  The probability of selecting exactly 1 defective fastener is

(one defective fastener)  

There is a 15.03% probability of randomly selecting exactly one defective fastener.

In this example, there are 120 fasteners in the shipment. Four fasteners in the shipment are defective. The remaining 116 fasteners are not defective. The quality-control manager randomly selects five fasteners.

To calculate the probability of selecting exactly one defective fastener, you must calculate

To calculate the numerator, enter the first value, 4, press MATH, highlight PRB and select 3:nCr and enter the next value, 1. Next press x and enter the next value, 116, press MATH, highlight PRB and select 3:nCr, enter the next value, 4. Next press and enter the first value in the denominator, 120, press MATH, highlight PRB and select 3:nCr, enter the next value, 5. Press ENTER and the answer will be displayed on your screen.