a very easy probability question...

[98.40_for_sure]

i just started working on the probability chapter and its doing my head in
i cant think of how to do this!

A die is weighted as follows:
Pr(2) = Pr(3) = Pr(4) = Pr(5) = 0.2            Pr(1) = Pr(6) = 0.1
The die is rolled twice and the smaller of hte numbers showing is noted. Let Y represent this value.
Find Pr(Y=1)

this is the second exercise of the chapter so please dont solve using some complex methods
as i assume the question is asking for some sort of easy way to solve it

[Russ]

It's asking what the probability is of a "1" being rolled on one die but not on both. If Pr(Y=1) = .1, then Pr(Y=/=1) = .9

So .1*.9 = .09

[98.40_for_sure]

thats not the answer though :S

[Russ]

Well, that's what I get for trying to answer maths questions lol. Hopefully someone else can help you!

[98.40_for_sure]

hahah thanks man

[QuantumJG]

It's asking what the probability is of a "1" being rolled on one die but not on both. If Pr(Y=1) = .1, then Pr(Y=/=1) = .9

So .1*.9 = .09

I think this is wrong.

We want to find the probability that the number will be 1.

So we could roll:

(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(3,1),(4,1),(5,1),(6,1)

Pr(Y=1) = 0.1*0.1 + 0.1*0.2 + 0.1*0.2 + 0.1*0.2 + 0.1*0.2 + 0.1*0.1 + 0.2*0.1 + 0.2*0.1 + 0.2*0.1 + 0.2*0.1 + 0.1*0.1

So Pr(Y=1) = 0.19

Just check my arithmetic. 

[vexx]

i just started working on the probability chapter and its doing my head in
i cant think of how to do this!

A die is weighted as follows:
Pr(2) = Pr(3) = Pr(4) = Pr(5) = 0.2            Pr(1) = Pr(6) = 0.1
The die is rolled twice and the smaller of hte numbers showing is noted. Let Y represent this value.
Find Pr(Y=1)

this is the second exercise of the chapter so please dont solve using some complex methods
as i assume the question is asking for some sort of easy way to solve it

well the dice is rolled twice, so there are 36 possibilities. and since we are looking for only when the smallest number is 1, we want any two rolls with a 1 in it, which is 1,1;1,2;1,3..... and pretty sure there is a 11 since there are 1 x 12 - 1 = 11 (cannot have 1,1 twice).. and the ones of 1,1 1,6 and 6,1 are 0.1*0.1 + ... + ... there are 3 so = 0.03
and there are 11-3 = 8 more, which are 0.02*0.01 probability each added to be 0.16
and then add them, total probability = 0.03+.16 = 0.19
hope this is correct:)

[98.40_for_sure]

umm... the worked solutions say Pr(Y=1) = 0.1 + 0.1 - 0.1 * 0.1 = 0.19
why is it so simple yet so hard!! D:

[98.40_for_sure]

ahh vexx got the answer, but i dont get the method worked solutions used :S

[vexx]

ahh vexx got the answer, but i dont get the method worked solutions used :S

yeah mine took a bit longer perhaps, but definitely easier when spread it out like i did.

[98.40_for_sure]

i get it now! thanks heaps vexx!

[m@tty]

First they add the probability that each of the die are 1, then subtract the probability of both of them being 1.