Conditional Probability?

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chetpuff

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This isn't a homework question but I'm trying to help a friend figure it out. If you have an 85% chance of Event A happening, but if it doesn't happen you have a 90% chance of it happening the second time you try it, what is the probability of Event A happening within two tries?
It's been a while since I took statistics and I'm not sure the right way to go about it. The only thing I can think of is 85% * 90% but I'm pretty sure that's not right.
 
chetpuff said:
This isn't a homework question but I'm trying to help a friend figure it out. If you have an 85% chance of Event A happening, but if it doesn't happen you have a 90% chance of it happening the second time you try it, what is the probability of Event A happening within two tries?
It's been a while since I took statistics and I'm not sure the right way to go about it. The only thing I can think of is 85% * 90% but I'm pretty sure that's not right.
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Suppose that [imath]A_n[/imath] is the event that event [imath]A[/imath] occurs on the [imath]n^{th}[/imath] trial.
You are asked for [imath]\mathcal{P}(A_1\vee A_1^cA_2)[/imath].
That is [imath]\mathcal{P}(A_1)+\mathcal{P}(A_1^c\cap A_2)=\mathcal{P}(A_1)+\mathcal{P}(A_2|A_1^c)\mathcal{P}(A_1^c)=0.85+(0.9)(0.15)[/imath]

[imath][/imath][imath][/imath][imath][/imath]
 
pka said:
Suppose that [imath]A_n[/imath] is the event that event [imath]A[/imath] occurs on the [imath]n^{th}[/imath] trial.
You are asked for [imath]\mathcal{P}(A_1\vee A_1^cA_2)[/imath].
That is [imath]\mathcal{P}(A_1)+\mathcal{P}(A_1^c\cap A_2)=\mathcal{P}(A_1)+\mathcal{P}(A_2|A_1^c)\mathcal{P}(A_1^c)=0.85+(0.9)(0.15)[/imath]

[imath][/imath][imath][/imath][imath][/imath]
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What about [imath]\Pr(A_1 \cap A_2)?[/imath]
 
BigBeachBanana said:
What about [imath]\Pr(A_1 \cap A_2)?[/imath]
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Well what about it? [imath]\mathcal{P}(A_1 \cap A_2)=\mathcal{P}(A_2|A_1)\mathcal{P}(A_1)[/imath]
But that was not the question, which was: what is the probability of Event A happening within two tries?
the event [imath]A[/imath] happens the first time or not the first tine but does happen the second time.[imath][/imath][imath][/imath]
 
pka said:
Well what about it? [imath]\mathcal{P}(A_1 \cap A_2)=\mathcal{P}(A_2|A_1)\mathcal{P}(A_1)[/imath]
But that was not the question, which was: what is the probability of Event A happening within two tries?
the event [imath]A[/imath] happens the first time or not the first tine but does happen the second time.[imath][/imath][imath][/imath]
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Why can't event A occur both times? Your answer assumes that the second attempt only occurs if the first has failed.
 
BigBeachBanana said:
Why can't event A occur both times? Your answer assumes that the second attempt only occurs if the first has failed.
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Yes indeed you are correct. That is exactly what the question asks for: on the first or on the second.
Why in world would we have a second try if it happens on the first try? The thread's title is: Conditional Probability.
 
blamocur said:
As I understand it, we don't care about [imath]Pr(A_1 \cap A_2)[/imath], but only about the total [imath]Pr(A_1) = \Pr(A_1 \cap A_2) + Pr(A_1 \cap A^c_2)[/imath]
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chetpuff said:
This isn't a homework question but I'm trying to help a friend figure it out. If you have an 85% chance of Event A happening, but if it doesn't happen you have a 90% chance of it happening the second time you try it, what is the probability of Event A happening within two tries?
It's been a while since I took statistics and I'm not sure the right way to go about it. The only thing I can think of is 85% * 90% but I'm pretty sure that's not right.
Click to expand...
Did you miss read the question?
 
pka said:
Why in world would we have a second try if it happens on the first try? The thread's title is: Conditional Probability.
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Suppose A is the event of getting an Ace from a deck of cards. As there a multiple aces in the deck, it is possible to get an Ace on the first, and on the second attempt. The phrase "conditional probability" doesn't imply continuing until success.
 
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blamocur said:
As I understand it, we don't care about [imath]Pr(A_1 \cap A_2)[/imath], but only about the total [imath]Pr(A_1) = \Pr(A_1 \cap A_2) + Pr(A_1 \cap A^c_2)[/imath]
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To me, "what's the probability of Event A happening within two tries?" means the probability of getting an A on the first try or getting an A on the second try, or getting A on both tries. But that's up to the OP to clarify.
 
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BigBeachBanana said:
To me, "what's the probability of Event A happening within two tries?" means the probability of getting an A on the first try or getting an A on the second try, or getting A on both tries. But that's up to the OP to clarify.
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I agree, and [imath]Pr(A_1)[/imath] includes the first and the third probabilities from your list.
 
BigBeachBanana said:
To me, "what's the probability of Event A happening within two tries?" means the probability of getting an A on the first try or getting an A on the second try, or getting A on both tries. But that's up to the OP to clarify.
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But getting A on the first try includes getting A on both! (Unless you mean, only on the first try.) So even if you make both tries regardless of whether the first succeeded, you still want [imath]\mathcal{P}(A_1)+\mathcal{P}(A_1^c\cap A_2)[/imath], which is the same as [imath]\mathcal{P}(A_1\cap A_2^c)+\mathcal{P}(A_1\cap A_2)+\mathcal{P}(A_1^c\cap A_2)[/imath].

You could also work this out by finding the probability of not getting A at all.
 
Dr.Peterson said:
But getting A on the first try includes getting A on both! (Unless you mean, only on the first try.) So even if you make both tries regardless of whether the first succeeded, you still want [imath]\mathcal{P}(A_1)+\mathcal{P}(A_1^c\cap A_2)[/imath], which is the same as [imath]\mathcal{P}(A_1\cap A_2^c)+\mathcal{P}(A_1\cap A_2)+\mathcal{P}(A_1^c\cap A_2)[/imath].

You could also work this out by finding the probability of not getting A at all.
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Yes, I realized that now.

For the record, @pka 's answer stands correct, but I had trouble with his reasoning.
[math]\Pr(\text{Event A happening within two tries})= \Pr(A_1 \cap A_2)+\Pr(A_1 \cap A_2^c) +\Pr(A_1^c \cap A_2)\\= \Pr(A_1)+\Pr(A_1^c \cap A_2)=\Pr(A_1)+\Pr(A_2|A_1^c)\Pr(A_1^c)\\ =0.85+0.9(0.15)=0.985[/math]
 
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[imath] {\rm P}({A_1} \cup {A_2}) = {\rm P}({A_1}) + {\rm P}({A_2}) - {\rm P}({A_1} \cap {A_2}) \\ {\rm P}({A_1} \cup {A_2}) = {\rm P}({A_1}) + \left[ {{\rm P}({A_2} \cap {A_1}) + {\rm P}({A_2} \cap {A_1}^c)} \right] - {\rm P}({A_1} \cap {A_2}) \\ {\rm P}({A_1} \cup {A_2}) = {\rm P}({A_1}) + {\rm P}({A_2} \cap {A_1}^c) \\ {\rm P}({A_1} \cup {A_2}) = {\rm P}({A_1}) + {\rm P}({A_2}|{A_1}^c){\rm P}({A_1}^c) \\ [/imath]
 
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