Question

17) You are playing in a lottery where 3 balls are picked to determine the winner. There are a total of 30 balls and each ball is labeled with only one number from 1-30. The balls are not returned after they are drawn. a. What is the probability of the first ball being labeled with an odd number? b. Suppose you bought one ticket, what is the probability that if you chose 8-12-21 as your numbers that you will win? c. Suppose before the drawing you decided to buy another ticket in addition to the ticket you already have; you chose the numbers 10-19-29. What is the probability that you will win now?

Answer 1

We know that there are 30 balls numbered from 1 to 30.

a.

The number of odd balls are 15.

The total number of balls are 30.

Thus, the probability that the first ball is odd is 15/30 = 0.5

b.

The first number can be either 8,12 or 21.

The second number can be any of the other two numbers.

The third number has to the left over number.

Thus, the probability that the first number is 8,12 or 21 is 3/30.

The probability that the second number is the other 2 numbers is 2/29.

The probability that the third number is the last number is 1/28.

Thus, the probability of winning is 3/30 * 2/29 * 1/28.

= 0.0002463054

c.

Since all the three numbers are different the probability of winning with the new ticket is the same as the old ticket.

The probability we may win now= Probability of winning using the old ticket + probability of winning using new ticket

= 0.0002463054+0.0002463054

= 0.0004926108