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 the space - shuttle system

The system fails only when all three parts (A, B, and C) fail. The probability that part A fails is $P(A_{fail})=1 - 0.95=0.05$, the probability that part B fails is $P(B_{fail})=1 - 0.90 = 0.10$, and the probability that part C fails is $P(C_{fail})=1 - 0.80=0.20$.

Step 2: Calculate the probability of system failure

Since the failures of parts A, B, and C are independent events (in the context of the redundancy design), the probability of the entire system failing is the product of the probabilities of each part failing. So, $P(system_{fail})=P(A_{fail})\times P(B_{fail})\times P(C_{fail})$.

Step 3: Substitute the values

$P(system_{fail})=0.05\times0.10\times0.20$.
$P(system_{fail}) = 0.001$.

Answer:

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