Question

Two players, Hazel and Ross, play a game with a biased dice and a fair dice. Hazel chooses one of the two dice at random and rolls it. If the score is 5 or 6 she wins a point.

Hazel chooses a dice, rolls it and wins a point. Find the probability that she chose the biased dice.
1=1/18
2=1/9
3=1/9
4=1/9
5=1/9
6=1/2

Answer 1

See my response from earlier for help http://openstudy.com/users/order#/updates/4ee31fade4b0a50f5c5669af

Answer 2

I saw that one... Am not sure how to use it to answer this question though.

Answer 3

dalvoron can u hlp me after this?

Answer 4

Hrm, it is a bit tougher alright. I think what you need to do is compare the P(5,6|biased) and P(5,6|unbiased). Their sum have to be 1 in this case, since we already know she won the round (i.e. rolled a 5 or a 6).

Answer 5

Ok, I'll try that out and tell you if I get the answer.

Answer 6

dalvoron??

Answer 7

No, never mind, that's wrong.

Answer 8

I haven't gotten the answer... Yeh I thought so.

Answer 9

I'm completely lost, sorry!

Answer 10

Hmmm... Ok. Do you know anyone else who may be able to help?

Answer 11

It's hard to explain my logic here, but look at the ratio of getting a 5/6 on the two dice:
\[biased : unbiased = \frac{1}{9}+\frac{1}{2}:\frac{1}{6}+\frac{1}{6}\]\[=\frac{11}{18}:\frac{6}{18}\]\[=11:6\]So the odds of her having picked the biased die is 11:6, so the probability is \[\frac{11}{11+6}=\frac{11}{17}\]

Answer 12

I understand every part up to, Why do you add 11 to 6 in the denominator, if the ration is 11/6?

Answer 13

11:6 means that in 11 scenarios, A happens, and in 6 scenarios, B happens. How many scenarios are there in total? Well there's 11+6=17. What is the probability that A happens? Well it's 11/17.

Answer 14

It probably makes more sense in two steps like that.

Answer 15

It does make a bit more sense, but I don't think I fully grasped it. I understand the ratio concept, but I still don't understand the last step...

Answer 16

Here is a more traditional way to do it...

\[P(B|W)=\frac{P(B,W)}{P(W)}=\frac{P(W|B)P(B)}{P(W)}\]

\[=\frac{P(W|B)P(B)}{P((W\cap B)\cup (W\cap B'))}\]

\[=\frac{P(W|B)P(B)}{P(W\cap B)+P(W\cap B')}\]

\[=\frac{P(W|B)P(B)}{P(W|B)P(B)+P(W|B')P(B')}\]

\[=\frac{\frac{11}{18}\times\frac{1}{2}}{\frac{11}{18}\times\frac{1}{2}+\frac{2}{6}\times\frac{1}{2}}=\frac{11}{17}\]

Answer 17

Thanks, I understand it now :)