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7-4 - Chapter 7.4 Counting Techniques and Probability In...

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Chapter 7.4: Counting Techniques and Probability In many probability problems, the sample space S is too large to count simply by listing all sample points. If this is the case, we may use our different counting techniques (i.e. permutations, combinations, etc.) to find n ( S ) and n ( E ). Example: In the lottery BC-49, a player selects six different numbers from 1 , 2 , 3 ,..., 49. On the day of the lottery drawing, the lottery official randomly selects the six winning numbers. If the player has exactly three of these numbers, then the player wins a \$10 prize. What is the probability of a player winning the \$10 prize with exactly one ticket? Let E be the event that exactly three numbers match. We need to count n ( S ) and n ( E ). First, the sample space consists of all possible selections of six numbers. The total number of ways in which 6 numbers can be selected from 49 is parenleftbigg 49 6 parenrightbigg . Thus, n ( S ) = parenleftbigg 49 6 parenrightbigg . To count n ( E ), there are two tasks: selecting three of the six winning numbers (which can be done parenleftbigg 6 3 parenrightbigg ways) and selecting three of the 43 remaining non-winning numbers (which can be done parenleftbigg 43 3 parenrightbigg ways). Thus, n ( E ) = parenleftbigg 6 3 parenrightbiggparenleftbigg 43 3 parenrightbigg . This gives us: P ( E ) = n ( E ) n ( S ) = parenleftbigg 6 3 parenrightbiggparenleftbigg 43 3 parenrightbigg parenleftbigg 49 6 parenrightbigg 0 . 0176 The probability of winning the \$10 prize is approximately 0.0176. Example: If all six numbers match, then the player wins the grand prize (usually at least \$1 000 000). What is the probability that a player wins the grand prize with one ticket? Let E be the event that all six numbers match on the ticket. 1

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Since the experiment has not changed, the sample space is the same, and n ( S ) = parenleftbigg 49 6 parenrightbigg . To count n ( E ) there is only one task: out of six winning numbers, select all six. This can be done parenleftbigg 6 6 parenrightbigg ways. Thus, n ( E ) = parenleftbigg 6 6 parenrightbigg . This gives us: P ( E ) = n ( E ) n ( S ) = parenleftbigg 6 6 parenrightbigg
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