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Assuming you have an unfair die where the probabiity of rolling a certain number is defined as follows: $$ \begin{array}\\ \text{Die Face} & 1 &2&3&4&5&6\\ \hline\\ \text{Probability} & .05 &.2 & .1 &.2 &.3 & .15 \end{array} $$ Find the probability that we roll an even number followed by an odd number.

I would think this is a conditional probability question, however, in the solutions it simply said that rolling an odd number first and rolling an even number second are two events $A$ and $B$ respectively, and then by independence said: $P(A)=.05+.1+.3=.45$ and $P(B)=.2+.2+.15=.55$ and $P(A\cap B)=P(A)P(B)=(.45)(.55)=.248$. When I worked this out first I thought this would be done by conditional probability. Why would my reasoning be wrong here? And why can we solve this problem like this?

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Maybe you can provide the exact formulation of the problem? So far, I would agree it is ambiguous –  Ilya Jan 5 '13 at 14:47
    
@Ilya This is pretty close to what it is exactly, I'll change it a little. The solution i posted is exactly what the solutions say too... It gave the distribution and asked precisely to find the probabilities that we roll an even number following an odd number. –  TheHopefulActuary Jan 5 '13 at 15:10
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@Kyle, by the way note that product is commutative, so the probability of rolling odd followed by even equals that of rolling even followed by odd. –  alancalvitti Jan 5 '13 at 15:25
    
It is a conditional probability problem, but a specially easy one. By independence, $\Pr(B|A)=\Pr(B)$. –  André Nicolas Jan 5 '13 at 17:05
    
@AndréNicolas: but $P(B)$ seems to me different from the answer $.248 = P(A)P(B)$ –  Ilya Jan 6 '13 at 11:02
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2 Answers 2

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It’s not a problem in conditional probability: it simply asks for the probability of a specific sequence of independent events, which is of course the product of their individual probabilities, just as the text’s answer has it.

It’s no different in principle from the following question: given a fair coin, what is the probability of tossing a head followed by a tail?

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The content of your answer is fine, but it's incorrect to say that "it's not a problem in conditional probability" - since Bayes theorem is derived from the product rule. It just happens that $P(B|A)=P(B)$ in this case b/c of independence, as Ahshay writes below. –  alancalvitti Jan 5 '13 at 15:20
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Your guess that its a conditional probability question is correct. This question is based on joint / conditional probability. So if you take A-> event where odd number occurs at first throw and let B-> be event where even number occurs at second throw. So you need a joint probability P(A,B) which is given as P(A,B)=P(B/A)*P(A). But since B is independent of A (*That's because throwing die second time doesn't depends on result you get first time). Hence we get P(A,B)=P(B)*P(A). Then as you stated we get P(A) and P(B) since getting odd number 1,3,5 are independent of each other and getting even number 2,4,5 are independent of each other. Hence you get P(A) = 0.45 and P(B) = 0.55 and hence P(A,B) = 0.2475. Please feel free to ask if you have any question regarding this description.

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