Practice Problems

9 Practice Problems

1. In a reliability test, a light switch is turned on and off until it fails. If the probability that the switch will fail every flip is 0.001, what is the probability that the switch will fail after exactly 1200 flips?

2. Three people are selected at random. Assume there are 365 days in a year. (a) What is the probability that all three people have the same birthday? (b) What is the probability that none of the three have the same birthday?

3. Freddie picks a number in the from 1 to 2023. At the same time, Toby picks a number from 1 to 4046. Assume that they have picked different numbers. What is the probability that Toby has picked a number higher than Freddie's?

9.1 Solutions

1. In order for the light switch to fail after exactly 1200 flips, it must have functioned successfully 1200 times first, and then fail right afterwards. Thus, our answer is $(1 - 0.001)^{1200} \times 0.001$.

2. (a) Let's select any day X for the first person in our group's birthday. The probability that the second person's birthday is X is $\frac{1}{365}$, and the probability that the third person's birthday is X is also $\frac{1}{365}$. Thus, the total probability is $1 \times \frac{1}{365} \times \frac{1}{365} = \frac{1}{133225}$.

   (b) Once again, Let's select any day X for the first person in our group's birthday. The probability that the second person's birthday is not X (call it Y) is $\frac{364}{365}$, as there are 364 days left to choose from. Using similar logic the probability that the third person's birthday is unique is $\frac{363}{365}$, as there are 363 days to choose from that are not X or Y. This leaves our final answer as $1 \times \frac{364}{365} \times \frac{363}{365} = \frac{363 \times 364}{365^3} = \frac{132132}{133225}$.

3. There is a $\frac{1}{2}$ chance that Toby picks a number from 2024 to 4046. In this case, Toby will always have picked a number higher than Freddie, because the largest number Freddie can pick is 2023. The other half of the time, both Toby and Freddie have picked a number from 1 to 2023. In this case, each has a $\frac{1}{2}$ chance of picking a number higher than the other. Thus, the total probability that Toby has picked a higher number than Freddie is $\frac{1}{2} \times 1 + \frac{1}{2} \times \frac{1}{2} = \frac{3}{4}$