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SOURCE:COMPETITION Number of Problems: 14. FOR PRINT ::: (Book)
A bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out. What is the probability that all beads in the bag are red after three such replacements?
We can divide the case of all beads in the bag being red after three replacements into three cases.
The first case is in which the two green beads are the first two beads to be chosen. The probability for this is
The second case is in which the green beads are chosen first and third. The probability for this is
The third case is in which the green beads are chosen second and third. The probability for this is
Add all these cases together
One fair die has faces , , , , , and another has faces , , , , , . The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?
In order to obtain an odd sum, exactly one out of the two dice must have an odd number. We can easily find the total probability using casework.
Case 1: The first die is odd and the second die is even.
The probability of this happening is
Case 2: The first die is even and the second die is odd.
Adding these two probabilities will give us our final answer.
Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?
In order for the product of the numbers to be prime, of the dice have to be a , and the other die has to be a prime number. There are prime numbers (, , and ), and there is only one , and there are ways to choose which die will have the prime number, so the probability is .
An envelope contains eight bills: ones, fives, tens, and twenties. Two bills are drawn at random without replacement. What is the probability that their sum is $ or more?
The only way to get a total of $ or more is if you pick a twenty and another bill, or if you pick both tens. There are a total of ways to choose bills out of . There are ways to choose a twenty and some other non-twenty bill. There is way to choose both twenties, and also way to choose both tens. Adding these up, we find that there are a total of ways to attain a sum of or greater, so there is a total probability of .
Coin is flipped three times and coin is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?
There are 4 ways that the same number of heads will be obtained; 0, 1, 2, or 3 heads.
The probability of both getting 0 heads is .
The probability of both getting 1 head is
The probability of both getting 2 heads is
The probability of both getting 3 heads is
Therefore, the probabiliy of flipping the same number of heads is: