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Home > Algebra 2 > Chapter 12 > 12.5 Probability of Independent and Dependent Events > 12.5 Problem Solving Help
 
   
Return to book index Chapter 12 : Probability and Statistics
12.5 Problem Solving Help

Lesson 12.5: Help for Exercise 29 on page 735

You may want to rephrase the second part of this question as follows: "How many tickets would a person have to buy to have a 1% chance that at least one ticket wins." This takes into account the fact that sometimes more than 1 ticket may be a winner in any given trial. The words at least suggest that complementary probability can be used to solve the problem. The complement of having at least one winning ticket is having no winning tickets. Use the verbal model shown below to set up the problem:

P(having at least one winning ticket)
=0.01
1-P(having no winning tickets)
=0.01

To find the number of combinations of tickets, use the formula . There is a selection of 6 numbers taken from a group of 42 numbers. Only one of the combinations will be a winning ticket, so to find the probability of having no winning ticket, subtract 1 from the number of combinations and divide by the total number of combinations. When you find an algebraic expression to replace P(having no winning tickets), make sure that you include the variable n which represents the number of tickets that must be purchased. The variable n is the exponent of the probability of a single ticket not winning the lottery. You might want to refer to Example 4 on page 731 since this is a similar problem.