Twenty tires are tested to see if they last as long as the manufacturer claims they do. Three tires fail the test. Two tires are selected at random without replacement for an inspection.
Find the probability that both tires fail the test.
Which is greater, the probability that the first tire fails and the second one passes? or vice-versa?
p(f) = 3/20
p(first) = 3/20 = .15
p(second) = 2/19 = .105
p first and second = .15 * .105 = .0158
first fail, second pass
= 3/20 * 18/19
first pass second fail = 17/20 * 3/19
To find the probability that both tires fail the test, we need to consider the number of ways we can select two tires that fail out of the total number of possible pairs of tires.
Step 1: Find the total number of possible pairs that can be selected. Since there are 20 tires in total, the number of possible pairs is given by the combination formula C(20, 2) = 20! / (2! * (20-2)! ) = 190.
Step 2: Find the number of ways two failing tires can be selected. With three tires failing, we have C(3, 2) = 3! / (2! * (3-2)! ) = 3 ways to select two failing tires.
Step 3: Calculate the probability that both tires fail. The probability is given by the ratio of the number of favorable outcomes (two failing tires) to the number of possible outcomes (total pairs). So, the probability is 3 / 190 ≈ 0.0158.
To determine whether the probability of the first tire failing and the second one passing is greater than the vice versa scenario, we need to consider the number of ways each scenario can occur.
Scenario 1: First tire fails, second tire passes. With three failing tires, we have 3 ways to choose the first failing tire, and since only 17 passing tires are left, we have 17 ways to choose the second passing tire. So, there are 3 * 17 = 51 ways for this scenario to occur.
Scenario 2: First tire passes, second tire fails. Since we have 17 passing tires, there are 17 ways to choose the first passing tire. Then, with three failing tires remaining, we have 3 ways to choose the second failing tire. So, there are 17 * 3 = 51 ways for this scenario to occur as well.
Both scenarios have the same number of ways to occur, so the probability of the first tire failing and the second one passing is equal to the probability of the first tire passing and the second one failing.