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.

calculate the probability that Hazel wins a point.

Biased Dice:

1=1/18
2=1/9
3=1/9
4=1/9
5=1/9
6=1/2

Answer 1

\[P(5,6)=P(bias)*P(5,6|bias)+P(unbias)*P(5,6|unbias)\]

Answer 2

Can you explain it more elaborately... Not sure how to apply the formula to the question

Answer 3

The probability of getting a 5 or a 6 firstly depends on which dice you pick. You can pick the biased die, OR the unbiased die. That means those probabilities are added:\

The next set of probabilities is about whether you pick a 5 or a 6. The probability is dependent on which die you picked at first. You get the biased die AND roll a number, OR get the unbiased die AND roll a number. P(A|B) means the probability, taking into account that B has already happened, so P(number|biased) means the probability of getting some number, taking into account that you picked the biased die.
\[P(number)=P(biased)*P(number|biased)+P(unbiased)*P(number|unbiased)\]
You're trying to get a 5 OR a 6, so you're adding probabilities. For the biased die, this probability is 1/9 + 1/2. The unbiased die has equal probabilities for both: 1/3 + 1/3.\[P(5,6)=P(biased)*P(5,6|biased)+P(unbiased)*P(5,6|unbiased)\]

Answer 4

I still can't seem to get the answer though...

Answer 5

Well let's work it out bit by bit.
\(P(biased)=P(unbiased)=\frac{1}{2}\), since there's an equal chance of picking the biased or unbiased die.
\(P(5,6|biased)=\frac{1}{9}+\frac{1}{2}=\frac{11}{18}\)
\(P(5,6|unbiased)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}\)
\[P(5,6)=P(biased)*P(5,6|biased)+P(unbiased)*P(5,6|unbiased)\]\[P(5,6)=\frac{1}{2}*\frac{11}{18}+\frac{1}{2}*\frac{1}{3}\]

Answer 6

Yes, that makes a lot more sense! I just got the answer myself. I used 1/2, but I had no idea why I used it, and now I do. Thanks for your help!

Answer 7

You're welcome.