Amy S.
asked • 01/27/18Easy question however, I somehow can't do it
a pupil travels to school either by MRT or by bus. Th probability that the pupil will be late for school is 1/12 if he travels by MRT and 1/9 if he travels by bus.
a) if he travels by bus on two consecutive days, find the probability that he will be late for school on one out of the 2 days. Give your answer as a fraction reduced to its simplest form.
b) if he is equally likely to travel by MRT or by bus, calculate the probability that he will not be late for school on any given day
c) what is the probability that he will travel by MRT in two successive days and he will not be late on both days? (Given that he is equally likely to travel by a MRT or by bus)
May I have step by step workings please? I really don't understand how to do this question!
a) if he travels by bus on two consecutive days, find the probability that he will be late for school on one out of the 2 days. Give your answer as a fraction reduced to its simplest form.
b) if he is equally likely to travel by MRT or by bus, calculate the probability that he will not be late for school on any given day
c) what is the probability that he will travel by MRT in two successive days and he will not be late on both days? (Given that he is equally likely to travel by a MRT or by bus)
May I have step by step workings please? I really don't understand how to do this question!
More
1 Expert Answer
John M. answered • 01/27/18
Tutor
5
(79)
Engineering manager professional, proficient in all levels of Math
(a) P[late for school if traveling by bus] = 1/9
The question appears to be asking for the probability of being late on exactly one of the two school days.
You can look at this in the following way: P[late exactly one day] = 1- P[not being late either day] - P[being late both days]
P[not being late either day] = 8/9 * 8/9 = 64/81
P[being late both days] = 1/9 * 1/9 = 1/81
P[late exactly one day] = 1- 64/81 - 1/81 = 16/81
Another way to solve this problem is to say that the P[late exactly one day] = P[late on day 1]*P[not being late on day 2] + P[not late on day 1] * P[late on day 2]
And since the Probability of being late is 1/9, you can substitute the values to arrive at: (1/9 * 8/9) + (8/9 * 1/9) = 8/81 + 8/81 = 16/81
Using either method, you arrive at the same answer for (a). The answer is 16/81
Now for part (b)
P[traveling by bus] = 1/2
P[traveling by MRT] = 1/2
P[being late if traveling by bus] = 1/9
P[being late if traveling by MRT] = 1/12
So P[late on any given day] = (1/2 * 1/9) + (1/2 * 1/12) = 1/18 + 1/24.
Common denominator is 72. So, 1/18 + 1/24 = 4/72 + 3/72 = 7/72
And P[not late on any given day] = 1- P[late on any given day] = 1 - 7/72 = 65/72
Answer to part (b) = 65/72
John M.
For part (c), you must first notice that the two events are "independent" of one another. In other words, being late on the first day is independent of being late on the second day.
P[not being late on both days] = P[not being late on day 1] * P[not being late on day]
P[not being late on day 1] = 1/2 * 11/12 = 11/24
P[not being late on day 2] = 1/2/* 11/12 = 11/24
Notice that we multiplied each equation above by 1/2 because that is the probability of traveling by MRT on any given day.
P[not being late on both days] = 11/24 * 11/24 = 121/576
Answer to part (c) = 1/576
It may help you to understand this type of problem if you consider all the things that could happen in 2 successive days. The pupil could:
1) take the MRT on day 1 and the MRT on day 2 (1/4 probability)
2) take the MRT on day 1 and the bus on day 2 (1/4 probability)
3) take the bus on day 1 and the bus on day 2 (1/4 probability)
4) take the bus on day 1 and the MRT on day 2 (1/4 probabilty)
Since these are the only possibilities, the sum is 4 * 1/4 = 1.0.
Now there are also 4 possibilities of what can happen in terms of arrival time
5) not late on day 1 and not late on day 2
6) not late on day 1 and late on day 2
7) late on day 1 and not late on day 2
8) late on day 1 and late on day 2
The probabilities of the arrival time depends on whether the pupil travels by bus or MRT. If traveling by MRT, then:
5M) 11/12 * 11/12 = 121/144
6M) 11/12 * 1/12 = 11/144
7M) 1/12 * 11/12 = 11/144
8M) 1/12 * 1/12 = 1/144
Again, if you add 5M thru 8M, you will arrive at a total of 144/144 = 1.0, because these are all the possibilities if a pupil travels by MRT on two consecutive days.
You could do a similar set of equations if traveling by bus, and call these probabilities 5B thru 8M.
Getting back to part (c), you can use the above probabilities to find the answer:
P[not being late on both days when taking MRT] = P[1] * P[5M] = 1/4 * 121/144 = 121/576
Report
01/27/18
Amy S.
Hi, I don't understand this part.
P[not late on any given day] = (1/2 * 1/9) + (1/2 * 1/12) = 1/18 + 1/24.
Isn't 1/9 and 1/12 the probabilities of being late if he travels by bus or MRT respectively? Hence, why are the probabilities for latecoming used in the above working to find the probability of not being late?
Report
01/29/18
John M.
Amy: You are correct. My original answer calculated the probability of being late. I should have subtracted this answer from 1.0. I edited the reply to reflect this.
Alternately, the answer could have been found by P[not late on any given day] = (1/2 * 8/9) + (1/2 * 11/12) = 8/18 + 11/24 = 32/72 + 33/72 =
65/72
Report
01/30/18
Amy S.
Wow. Thank you so so much for the awesome and detailed explanation. I truly appreciate it and now it's so clear!
Report
02/01/18
John M.
You're welcome!
Report
02/01/18
Still looking for help? Get the right answer, fast.
Ask a question for free
Get a free answer to a quick problem.
Most questions answered within 4 hours.
OR
Find an Online Tutor Now
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.
Amy S.
01/27/18