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The probability that there is no accident at a certain busy intersection is 95% on any given day, independently of the other days.

(a) Find the probability that there will be no accidents at this intersection during the next 7 days.
i) 0.95
ii) 0.307
iii) 0.698
iv) 0.028

(b) Find the probability that next September there will be exactly 2 days with accidents.
i) 0.028
ii) 0.001
iii) 0.090
iv) 0.002

(c) Today was accident free. Find the probability that there is no accident during the next 4 days, but there is at least one by the end of the 10th day.
i) 0.547
ii) 0.076
iii) 0.453
iv) 0.028

User Rlatief
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1 Answer

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Final answer:

The probabilities for the three parts of the question are approximately 0.698 for no accidents over 7 days, 0.090 for exactly 2 accidents in September, and approximately 0.327 for no accidents during the next 4 days but at least one by the end of the 10th day. Option iii) is correct for both part A and B, but part C does not match any of the provided options.

Step-by-step explanation:

To solve the student's question, we will use the rules of probability for independent events. Let's address each part of the question:

Part A (Probability for 7 consecutive days)

The probability of no accidents at the intersection for 7 consecutive days is the product of the daily probabilities since the days are independent events. Therefore, the calculation is:

(0.95)^7 = 0.6983

The probability is approximately 0.698 which corresponds to answer iii).

Part B (Exactly 2 days with accidents in September)

September typically has 30 days. We use the binomial probability formula for exactly 2 days with accidents, which is:

P(X = 2) = C(30, 2) * (0.05)^2 * (0.95)^28

After calculating, we get an approximate probability of 0.090, which corresponds to answer iii).

Part C (No accident for next 4 days but at least one by the 10th day)

The probability of having no accidents for the next 4 days is simply:

(0.95)^4 = 0.8145

To ensure at least one accident by the 10th day, we subtract the probability of no accidents for 10 days from 1:

1 - (0.95)^10 = 0.4013

Now, multiplying these probabilities gives us the final answer:


(0.8145) * (0.4013) = 0.327

However, this scenario is not represented by the provided options, indicating a possible error in the question or the need for further clarification.

User Guinevere
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