Kenneth S. answered • 01/26/18
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A) (1/9)2
...is probability of being late both days; P(on time both days) = the square of (8/9). Call these two results a and b, respectively. The P(being late exactly one day is 1 - a - b.
B) (½(11/12) + ½(8/9) ... FACTOR ONE HALF BECAUSE EQUALLY LIKELY TO CHOOSE MRT OR BUS; ASSOCIATED FACTORS ARE 1 - P(late) in each instance.
Kenneth S.
For b) I get 65/72 = 0.902777... which seems reasonable (the two probabilities for lateness being 0.08333 and 0.111... and that's an average 7/72 or 0.097 (this is just a crude check).
I may look at c) when I clear my desks of other business.
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01/26/18
Kenneth S.
Note that I have modified the answer for a).
For c), we'll assume MRT transportation used on two consecutive days. The question asks for probability of not being late on both days. This means, to me, that he could be on time both days, or late on exactly one of the two days.
We should use Binomial Distribution: (late + not.late)2 = (1/12)2 + 2(1/12)(11/12) + (11/12)2 is the expansion; first term is probability of on time both days; last term is probability of late both days; middle term is probability of one late, one on time.
Therefore P(not late on 2 consec. days, by MRT) = the sum of the first two terms above; this is 1 - (11/12)2.
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01/26/18
Amy S.
so your answer for part a) is (1/9)^2?
For part c, you mentioned the first term was the probability of on time both days which is (1/12)^2. But isn't 1/12 the probability that the pupil will be late if he travels by MRT?
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01/26/18
Kenneth S.
My (final) ANSWER for a) is NOT 1/81. Calculate a & b as I noted, then 1-a-b.
c) The binomial expansion is typed (laboriously) is correct. It represents probability of being late (by MRT) both days--you correctly detected a mistake in my typing (I said on time rather than late! With that corrected verbiage, I think the problem is correctly explained.
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01/27/18
Amy S.
01/26/18