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For every 10,000 miles driven, the probability that a school bus in the United States will be in at least one accident is 1/6. For 12 buses in the lot, what are the probabilities that:

assume that each is driven 10,000 miles.

So I think that I have to use a binomial distribution with p=1/6, and q=1-p=5/6 and n=12, thus:

a. none will be in an accident $$P(0 accidents) = P(X=0) = \binom{12}{0}(1/6)^0(5/6)^{12} = (1/6)^{12}= 4.5939 x e^{-10}$$

b. at most 1 of the buses will have at least 1 accident $$P(≤1 accidents) = \binom{12}{1}(1/6)^1(5/6)^{11}+\binom{12}{0}(1/6)^0(5/6)^{12}= \frac{5^{11}}{6^{12}}(17)=0.3813$$

c. at least 2 buses will have accidents? $$P(at least 2 accidents) = 1-{\binom{12}{1}(1/6)^1(5/6)^{11}+\binom{12}{0}(1/6)^0(5/6)^{12}}= 1-{\frac{5^{11}}{6^{12}}(17)}=0.6187$$

P(≥2 accidents) = P(X≥2) = (12 2)(1/6)^2(5/6)^10 + (12 3)(1/6)^3(5/6)^9 + (12 4)(1/6)^4(5/6)^8 + (12 5)(1/6)^5(5/6)^7 + (12 6)(1/6)^6(5/6)^6 + (12 7)(1/6)^7(5/6)^5 + (12 8)(1/6)^8(5/6)^4 + (12 9)(1/6)^9(5/6)^3 + (12 10)(1/6)^10(5/6)^2 + (12 11)(1/6)^11(5/6)^1+ (12 12)(1/6)^12(5/6)^0

can someone please direct me to a page that can tell em how to format on this website? And is this the correct way of thinking or am i supposed to use a different type of distribution? To be honest I don't understand when one distribution is used vs another.

Math Major
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1 Answers1

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The event "at least $2$ buses have an accident" is the complement of the event "at most $1$ bus will have an accident." So the answer to c) is $1$ minus the answer to b).

The expression you obtained for c) is correct, but clearly far less computationally convenient. And a nuisance to type!

Remark: In calculating the answers to a) and b), a tacit independence assumption was made. This is not entirely plausible, for various reasons. If a bus driver has an accident, it may make the others (temporarily) more cautious. And the likelihood of accident may be affected by common external conditions, such as an ice storm.

With the assumption of independence, we have $12$ independent trials, in each of which the probability of "success" (!) is $\frac{1}{6}$. So the number of successes has binomial distribution with parameters $n=12$ and $p=\frac{1}{6}$.

André Nicolas
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