The probability of a randomly selected car crashing during a year in a certain country is .0479. If a family has three cars, find the probability that at least one of them has a car crash during a year.
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"At least one" = either 1, 2 or 3.
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
for 1: .0479 * (1-.0479)^2 = ?
for 2: .0479^2 * (1-.0479) = ?
for 3: .0479^3 = ?
Either-or probabilities are found by adding the individual probabilities.
To find the probability that at least one of the three cars has a car crash during a year, we can use the complement rule.
The complement rule states that the probability of an event occurring is equal to 1 minus the probability of the event not occurring.
In this case, we want to find the probability that none of the three cars have a car crash during a year. The probability of one car not crashing in a year is 1 - 0.0479 = 0.9521.
Since the cars are independent, we can multiply the probabilities of each car not crashing together: 0.9521 * 0.9521 * 0.9521 = 0.8575.
Finally, we can find the probability that at least one of the cars has a crash by subtracting the probability that none of them have a crash from 1: 1 - 0.8575 = 0.1425.
Therefore, the probability that at least one of the three cars has a car crash during a year is 0.1425, or 14.25%.
To find the probability that at least one of the three cars has a car crash during a year, we can use the concept of complementary probability. The complementary probability of an event A is equal to 1 minus the probability of the event not occurring (denoted as A').
In this case, let's find the probability that none of the three cars has a car crash during a year, denoted as event A.
The probability of a car not crashing during a year is 1 minus the probability of it crashing, which is 1 - 0.0479 = 0.9521.
Since each car is independent of the others, we can multiply the probabilities together to find the probability that none of the three cars has a car crash during a year: P(A) = 0.9521 * 0.9521 * 0.9521 = 0.8577.
Now, the probability of at least one of the three cars having a car crash during a year (event A') is 1 - P(A).
P(A') = 1 - 0.8577 = 0.1423.
Therefore, the probability that at least one of the three cars has a car crash during a year is 0.1423, which is approximately 14.23%.