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We have four coins in a box, let $p_i$ describe the probability of getting head from coin $i$: $p_1=0, p_2=0.25, p_3=0.5, p_4=0.75$.

We toss take one coin and toss it until we get head, what is the probability of tossing it exactly 3 times?

I can calculate it fairly easily with a tree to get: $0.25(0.25\cdot0.75^2+0.5^3+0.25^2\cdot0.75)=0.078125$

But why we don't have to divide that result by $|\Omega|=2^3$ (heads or tails 3 times) or by the probability of getting tails in the first two tries?

I.e: $A= \text{heads in the third try} \\ B=\text{not head in the first and second tries}$

$P(A|B)=\frac {P(A\cap B)}{P(B)}=\frac{0.078125}{0.25(0.75^2+0.5^2+0.25^2)}=0.357143$

I'm basically asking why is the answer $P(A \cap B)$ and not $P(A|B)$ or $P(A \cap B)/|\Omega|$ ?

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    $\begingroup$ You ask "why not...?" What would be your answer if the question"why...?" was posed? I really don't understand you. $\endgroup$
    – drhab
    Apr 20, 2015 at 15:44
  • $\begingroup$ The probability that the question asks and the conditional probability are different. "We take one coin and toss it until we get head" is a setting, the rule of the game, but not a given event. $\endgroup$
    – James Pak
    Apr 20, 2015 at 15:54
  • $\begingroup$ Because you're computing a joint probability (tails/tails AND heads), not a conditional probability (heads, given tails/tails). $\endgroup$
    – Brian Tung
    Apr 20, 2015 at 15:54
  • $\begingroup$ @drhab I'm asking why don't have to use the conditional probability formula or divide by $\Omega$ $\endgroup$
    – shinzou
    Apr 20, 2015 at 16:09
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    $\begingroup$ No, the question asks for the probability that you toss the coin exactly three times. Roughly speaking, it is assumed that we ask the question before we begin the game. If you had asked the question after you had tossed the coin twice (and gotten tails both times), then your answer of about $0.357$ would be correct. You would have been GIVEN the first two tosses as tails. But why would we assume that you ask the question only after the first two tosses have already happened? $\endgroup$
    – Brian Tung
    Apr 20, 2015 at 16:30

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You have written down two quantities: The first quantity 0.078... looks like the correct answer, which symbolically is $P(A \cap B)$. The second quantity $P(A \, | \, B)$ is the conditional probability where you have ignored (divided by) the initial probability $P(B)$ of not getting heads the first and second time. So that is a different quantity and not the answer to your question.

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  • $\begingroup$ Why is the answer $P(A \cap B)$ and not $P(A|B)$? $\endgroup$
    – shinzou
    Apr 20, 2015 at 16:10

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