The Student Room Group

Conditional probability

A bus serving a number of villages is due to arrive in a particular village at 10 o'clock. Past experience tells the people waiting in the village for the bus that the probability of the service being cancelled on any day is 0.05, and that, when it runs, the probability of the bus being late is 0.1 Draw a tree diagram to show this information.
Using your tree diagram, find
i)the probability that the bus has not arrived in the village at 10 o'clock,
ii)the conditional probability that the service has been cancelled, given that the bus has not arrived in the village at 10 o'clock.

I am confused in part ii because if the bus service has been cancelled, how can it arrrive or not arrive in the village at 10 o'clock? I willshow my working for part i because I am not sure if I am right or not. Can someone please teach me? I have exam in just two days.

i) P(bus has not arrived at 10 o'clock) = 0.95 * 0.1
(But I think it's wrong?)

I am completley confused at part ii, can someone please help? Thanks
Reply 1
i) P(Bus hasn't arrived at 10)= P(cancelled) or P(not cancelled and late)
= 0.05+(0.95x0.1)
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ii) You know the bus didn't arrive at 10 so what was the probability that the bus was cancelled?

Use the conditional probability P(AnB)= P(A|B)P(B)

So in this case, B is the event that the bus hasn't arrived at 10. and A is the event of the bus being cancelled.

So P(bus cancelled given that we know the bus didn't arrive at 10)= P(A|B)
=P(bus cancelled and doesn't arrive at 10) / P(bus hasn't arrived at 10)
=P(bus cancelled) / P(bus hasn't arrived at 10) [since if the bus is cancelled then it doesn't arrive]
=0.05/(0.05+0.95x0.1) [notice the denominator is the answer from (i)]
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probability isn't my strong point in maths, so anyone can feel free to correct me if I'm wrong.
Reply 2
Thnaks fo your help! I have 3 more questions, thanks!

1)Joseph and four friends each have an independent probability 0.45 of winning a prize. Find the probability that
a)exactly two of the friends win a prize(done)
b)Joseph and only one friend win a prize (ans = 0.135)

For part a is easy because it's like no restriction and so it's C(5, 2) * 0.45^2 * 0.55^3, but how about part b? How do I know that it must be Joseph and only one friend win a prize? :confused:

2)Show that, when two fair dice are thrown, the probability of obtaining a 'double' is 1/6, where a 'double' is defined as the same score on both dice. (done) Four players play a board game which requires them to take it in turns to throw two fair dice. Each player throws the two dice once in each round. When a 'double' is thrown the player moves forward six squares. Otherwise the player moves forward one square. Find
c)the probability that a 'double' occurs exactly once in 4 of the first 5 rounds. (ans. = 0.068)

3)The mass of grapes sold per day in a supermarket can be modelled by a normal distribution. It is found that, over a long period, the mean mass sold per day is 35.0kg, and that, on average, less than 15.0 kg are sold on one day in twenty.
a)Show that the standard deviation of the mass of grapes sold per day is 12.2 kg, correct to 3 significant figures.

How do I calculate the standard deviation without knowing the porbability? :confused:

Thanks for your help, I really appreciate it.
Reply 3
I would love to help you with these. But got to go. I've got lectures in 7 minutes.
Reply 4
Sorry to bump a very old thread but I'm trying to do this question myself now. When drawing the tree, which way is correct (see attachment)? I'm assuming its the first one, but this post has confused me:
Original post by mikeyT
i) P(Bus hasn't arrived at 10)= P(cancelled) or P(not cancelled and late)
= 0.05+(0.95x0.1)
--------------
ii) You know the bus didn't arrive at 10 so what was the probability that the bus was cancelled?

Use the conditional probability P(AnB)= P(A|B)P(B)

So in this case, B is the event that the bus hasn't arrived at 10. and A is the event of the bus being cancelled.

So P(bus cancelled given that we know the bus didn't arrive at 10)= P(A|B)
=P(bus cancelled and doesn't arrive at 10) / P(bus hasn't arrived at 10)
=P(bus cancelled) / P(bus hasn't arrived at 10) [since if the bus is cancelled then it doesn't arrive]
=0.05/(0.05+0.95x0.1) [notice the denominator is the answer from (i)]
--------------
probability isn't my strong point in maths, so anyone can feel free to correct me if I'm wrong.


Thanks

photo (4).jpg

EDIT: I don't know why the image has decided to rotate 180 on here... Also I'm only concerned by part ii
Reply 5
bump
Original post by danlocke
bump


Don't like either diagram.

The second is better though.

Here's my take on it. Where "Late" means "not arrived by 10:00" regardless whether it's running or not.
(edited 11 years ago)
Reply 7
Original post by ghostwalker
Don't like either diagram.

The second is better though.

Here's my take on it. Where "Late" means "not arrived by 10:00" regardless whether it's running or not.


Ok so am I right in thinking I want to work out P(C | L) = P(CnL)/P(L)?

I went with that and got P(L)=(0.05x1.0)+(0.95x0.1) = 0.145 and P(CnL)=(0.05x1.0) = 0.05

then did 0.05/0.145= 10/29

Is that right?
Original post by danlocke


then did 0.05/0.145= 10/29

Is that right?


Yep, I agree with that.
Reply 9
Original post by ghostwalker
Yep, I agree with that.


Great, thanks for your help :biggrin: