A consumer organization estimates that over a 1-year period 15 % of cars will need to be repaired once, 9 % will need repairs twice, and 4 % will require three or more repairs. If you own two cars, what is the probability that a) neither will need repair? b) both will need repair? c) at least one car will need repair?

Question

Mathematics

Braylon

4 June, 00:06

A consumer organization estimates that over a 1-year period 15 % of cars will need to be repaired once, 9 % will need repairs twice, and 4 % will require three or more repairs. If you own two cars, what is the probability that a) neither will need repair? b) both will need repair? c) at least one car will need repair?

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Answers (1)

Know the Answer?

Beckett Ortiz · Answer 1

a) The probability that neither will need repair is 51.84%

b) The probability that both will need repair is 7.84%

c) The probability that at least one car will need repair is 56%

Step-by-step explanation:

a) The probability that neither will need repair is 51.84%

(100 - 15 - 9 - 4) / 100 * (100 - 15 - 9 - 4) / 100

(72/100) * (72/100)

5,184 / 10,000 = 0.5184 in decimals and 51.84%

b) The probability that both will need repair is 7.84%

(15 + 9 + 4) / 100 * (15 + 9 + 4) / 100

(28/100) * (28/100)

784 / 10,000 = 0.0784 in decimals and 7.84%

c) The probability that at least one car will need repair is 56%

(15 + 9 + 4) / 100 + (15 + 9 + 4) / 100

(28/100) + (28/100)

56 / 100 = 0.56 in decimals and 56%

To calculate probabilities of two or more events are two rules: the sum and the product. The sum rule is used to calculate the probability of two or more events that cannot occur at the same time or are dependent. For example, the probability of having a 2 or a 3 when you throw a six-sided dice is 33%.

P (2) + P (3) = P

1/6 + 1/6 = 2/6 = 1/3 in fraction; 0.33 in decimals; 33% in percentage

On the other hand, the product rule is used to calculate the probability of two or more event that can occur at the same time or are independent. For example, the probability of having a 2 in both dices when you throw two six-sided dices is 2.78%.

P (2 in first dice) x P (2 in second dice) = P

1/6 * 1/6 = 1/36 in fraction; 0.0278 in decimals; 2.78% in percentage

Finally, you can use both rules. For example, the probability of having a 2 or a 3 in both dices when you throw two six-sided dices is 2.78%.

P (in first dice) x P (in second dice) = P

(1/6 + 1/6) * (1/6 + 1/6) = P

(2/6) * (2/6) = P

(1/3) * (1/3) = P

P = 1/9 in fraction; 0.111 in decimals; 11.11% in percentage