Probality question

Mark

On an unbiased die, the numbers 1, 3 and 4 are coloured red and the numbers 2, 5 and 6 are coloured black,

The die is thrown three times with the following outcome:

The second throw shows a red number and the sum of the numbers on the first and second throws is equal to the number on the third throw.

Find the probality of this outcome.

The answer is 10 over 216 (allegedly) which I've failed to get for several hours now. Halp meh.

silverside

there are 3 independent rolls, each of which can take values from 1 to 6. So the sample space consists of all ordered triplets from (1,1,1) to (6,6,6) which contains 6*6*6 or 216 equally likely events.

If you work out the ones that match your description, the only possible ones are
1,1,2
1,3,4
1,4,5
2,1,3
2,3,5
2,4,6
3,1,4
3,3,6
4,1,5
5,1,6

(Noting that a roll above 6 on the 3rd die is impossible)

so there are 10 events which make up the outcome.

Since there are 216 equally likely events in the sample space, and the probality of the sample space is 1, each event has probability 1/216. The probability of the outcome is the sum of the probability of individual events which make it up, or 10*1/216, = 10/216.

I know this is a bit wordy but you get the idea.

Mark

Thank you.

ApeXaviour

This problem has been giving me a lot of trouble... If you can help me out I'd appreciate it.


The probability that a certain flight departs on time is 0.8. The probability that it arrives on time is 0.9. Also, the probability that it both departs and arrives on time is 0.78. Find the probability that the flight:

(i) arrives on time given that it departed on time.

(ii) does not arrive on time given that it did not depart on time.



Cheers

smiles

P(A|B) is the probability that A occurs given that B occurs.

P(A|B) = P(B|A)P(A) / P(B)

P(A intersection = P(A)P(B|A) = P(B)P(A|B)

P(A|B) = P(A intersection / P(B)

Ok, so lets say:
A = departs on time
B = arrives on time.
¬A = doesn't depart on time
¬B = doesnt arrive on time.

P(¬A) = 1 - P(A)
P(¬B) = 1 - P(B)

We know that:
P(A) = 0.8
P(B) = 0.9
P(A intersection = 0.78

so
P(¬A) = 1 - 0.8 = 0.2 [ P(¬A) = 1 - P(A) ]
P(¬B) = 1 - 0.9 = 0.1 [ P(¬B) = 1 - P(B) ]

P(B|A) = 0.78 / 0.8 = 0.975 [ P(B|A) = P(A intersection / P(A) ]
P(A|B) = 0.78 / 0.9 = 0.86666... [ P(A|B) = P(A intersection / P(B) ]

Also note that:
P(A|B) + P(A|¬B) = P(A)
P(B|A) + P(B|¬A) = P(B)
Think about it in terms of Venn diagrams if you want, P(A|B) is one part of A... and so P(A|¬B) is the rest of it... add them together and you get the whole thing!

and
P(A|B) + P(¬A|B) = 1
P(B|A) + P(¬B|A) = 1
(basically the same as P(C) + P(¬C) = 1 - do the venn diagram thinking again!)

(i) arrives on time given that it departed on time.

so that is P(B|A) = 0.975 from above

(ii) does not arrive on time given that it did not depart on time.

so that is P(¬B|¬A)

P(¬B|A) = 1 - P(B|A) = 1 - 0.975 = 0.025

since P(B|A) + P(B|¬A) = P(B) then:
P(¬B|¬A) + P(¬B|¬¬A) = P(¬B)

ie. P(¬B|¬A) + P(¬B|A) = P(¬B) [ P(¬¬A) = P(A) ]
P(¬B|A) = 0.1 - 0.025 = 0.075

....... I think!!!!

<< Fio >>

silverside

Smiles, I don't think your answer is fully correct.

The way i did it was to draw a 2 X 2 Grid with 4 cells

DT,AT
DT, ANT
DNT,AT
DNT,ANT

where T stands for 'on time' and NT stands for 'not on time'.

Now you are given
DT,AT + DT,ANT = 0.8
DT,AT + DNT,AT = 0.9
DT,AT = 0.78
sum of all 4 cells = 1
so you can find

DT,AT = 0.78
DT, ANT = 0.02
DNT,AT = 0.12
DNT,ANT = 0.08

P(AT given DT) = 0.78 / (0.78 + 0.02) = 0.975

P(ANT given DNT) = 0.08 / (0.08 + 0.12) = 0.4

You have got Bayes' Rule a bit back to front. For a simple case like this it is easier to draw out a grid/venn diagram.

smiles

fair enough, what do you mean by "back to front" for the bayes theorem?

<< Fio >>

silverside

Actually you didnt use Bayes theorem (I think).

But you can't add conditional probabilities like you did.

i.e. it is not true that
P(A|B) + P(A|¬B) = P(A)

instead
P(A n + P(A n ~B) = P(A)