Question

1. a baseball team plays a best 2 of 3 game series. their chance of winning either of the first two games is 45%, but their chance of winning a 3rd game (if it occurs) is 80%. let x = the number of games won.
2. a soccer team plays two games. in each game, they have a 60% chance of winning, 30% chance of losing, and 10% chance of tying. let x = the number of games won.
3. space shuttle systems often were designed with triple redundancy (backup for the backup) to lower the chances of system failure as much as possible. part a is the main part, part b is the backup, part c is the backup’s backup. parts b and c sit dormant in the background unless the part in front of them fails. part a works 95% of the time, part b works 90% of the time, and part c works 80% of the time. draw a tree diagram for this situation, calculate the probability of each branch. what is the chance of a complete system failure?

Explanation:

Step 1: Analyze problem 7

We want to find the probability of system - failure. The system fails when part A fails, then part B fails, and then part C fails. Since these are independent events (when one part fails, the next - part's functionality is not affected in terms of its own probability), we use the multiplication rule for independent events.

Step 2: Calculate the probability of each part failing

The probability that part A fails is $P(A_{fail})=1 - 0.95=0.05$. The probability that part B fails is $P(B_{fail})=1 - 0.90 = 0.10$. The probability that part C fails is $P(C_{fail})=1 - 0.80=0.20$.

Step 3: Calculate the probability of system - failure

By the multiplication rule for independent events, the probability of the system failing is $P(system_{fail})=P(A_{fail})\times P(B_{fail})\times P(C_{fail})$.
$P(system_{fail})=0.05\times0.10\times0.20$
$P(system_{fail}) = 0.001$

Answer:

The probability of a complete system failure is $0.001$.