MA 111
Quiz 8—Section 4.2
February 28, 2008
1. (1 point) ID numbers consist of 6 digits, the first digit nonzero. Digits may be
repeated. What is the probability that an ID number selected at random ends in an even
number?
Total number of ID numbers: 9*10*10*10*10*10=900,000 by BCL.
Total number of ID numbers ending with an even digit: 9*10*10*10*10*5=450,000.
So the probability of selecting an ID number ending in an even digit is
450,000/900,000=1/2.
Note: You cannot use permutations or combinations here since choices may be repeated,
and permutations and combinations assume choices are not repeated.
2. (1 point) Gary plays Keno, and he chooses a 7-number card by choosing from the
numbers 1 to 80. The computer selects 20 numbers to be winning numbers. What is the
probability that Gary’s card has at least one winning number? (Hint: Use the
complement of this event.)
The complement of this event is that the card has no winning numbers.
The total number of cards is the total number of ways to select 7 different numbers out of
80 numbers, which is C(80,8).
The total number of cards with no winning numbers is the number of ways to select 7
losing numbers out of 80-20=60 losing numbers, which is C(60,7).
The probability that the card has no winning numbers is C(60,7)/C(80,7). Thus, the
probability that the card has at least one winning number is
1-C(60,7)/C(80,7)=.8784.
3. (2 points) Go back to Keno in number 2 above. What is the probability that Gary’s
card has exactly 4 winning numbers?
The total number of cards is C(80,7) as before.
The number of cards with 4 winning numbers is C(20,4)*C(60,3).
There are C(20,4) ways to choose the 4 winning numbers on the card.
There are C(60,3) ways to choose the 3 losing numbers on the card. So the total
number of these cards is the product of these two numbers.
So the probability that his card has 4 winning numbers on it is
(C(20,4)*C(60,3))/C(80,7)=.05219.