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Simple probability Q

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Two coins are tossed. What is the conditional probability that two heads result given that there is a t least one head?

let S be the sample space

S = {{HH},{HT},{TH,{TT}}

P(HH|HT,TH,HH) = P(HH n HT,TH,HH)/P(HT,TH,HH) = (1/4 * 3/4) / (3/4) = 1/4

is this correct? I'm finding it hard to determine what the question is actually asking
 

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  • #2
Ray Vickson
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Two coins are tossed. What is the conditional probability that two heads result given that there is a t least one head?

let S be the sample space

S = {{HH},{HT},{TH,{TT}}

P(HH|HT,TH,HH) = P(HH n HT,TH,HH)/P(HT,TH,HH) = (1/4 * 3/4) / (3/4) = 1/4

is this correct? I'm finding it hard to determine what the question is actually asking
Think about what the question says: what is the conditional probability of two heads if you are told you get at least one head? The information you are told eliminates one or more of the original sample points; can you say which sample point (or points) is (or are) eliminated? For the points that remain, what are the new, adjusted, probabilities?
 
  • #3
phinds
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"conditional" means what it says --- a condition has been imposed. So here are two different questions:

What are the odds that two coin tosses in a row will result in two heads?

What are the odds that two coin tosses in a row will result in two heads if you add the condition that it is known that one of the tosses results in a head?
 
  • #4
HallsofIvy
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{{H,H}, {H, T}, {T, H}, {T,T}} is the sample space for "flip two coins and observe which side is up". The condition, "at least one head" reduces that to {{H, H}, {H, T}, {T, H}}. What fraction of that is "two heads"?
 
  • #5
haruspex
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P(HH n HT,TH,HH)/P(HT,TH,HH) = (1/4 * 3/4) / (3/4)
You appear to have evaluated P(HH n HT,TH,HH) as P(HH)*P(HT,TH,HH). The two are not independent, so you cannot multiply their probabilities. Indeed, one is a subset of the other, so P(HH n HT,TH,HH) = P(HH).
 
  • #6
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{{H,H}, {H, T}, {T, H}, {T,T}} is the sample space for "flip two coins and observe which side is up". The condition, "at least one head" reduces that to {{H, H}, {H, T}, {T, H}}. What fraction of that is "two heads"?
so it's 1/3?
 
  • #7
phinds
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That's not what I get.
here is my working:

P(HH|HT,TH,HH) = P(HH)/P(HT,TH,HH) = (1/4)/(3/4) = 1/3
 
  • #9
phinds
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Here's my working for coin 1 / coin 2

H/H
H/T
T/H
H/H

the two HH's are NOT the same. They are the same only if you are looking for COMBINATIONS, not probabilities.
 
  • #10
Ray Vickson
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Here's my working for coin 1 / coin 2

H/H
H/T
T/H
H/H

the two HH's are NOT the same. They are the same only if you are looking for COMBINATIONS, not probabilities.
Every probability book I have ever seen would say there is just one HH if the coins are undistinguished (i.e., maybe distinguishable, but we don't bother to do it). You *would* be correct if you distinguished between the two coins, say one is red and the other is blue. Then the sample space would be {(RH,BH),(BH,RH),(RH,BT),(BH,RT),(RT,BH),(BT,RH),(BT,RT),(RT,BT)}, with (presumably) each sample point (i,j) having probability 1/8. One could then work the problem for this representation, but the final answer would be the same as in the simpler, undistinguished case.
 
  • #11
phinds
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Every probability book I have ever seen would say there is just one HH if the coins are undistinguished (i.e., maybe distinguishable, but we don't bother to do it). You *would* be correct if you distinguished between the two coins, say one is red and the other is blue.
But you CAN, and I believe MUST, distinguish between the coins. One is tossed first and the other is tossed second, and this gives 4 outcomes, as I stated.
 
  • #12
Ray Vickson
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But you CAN, and I believe MUST, distinguish between the coins. One is tossed first and the other is tossed second, and this gives 4 outcomes, as I stated.
But that is related to what I said in my post. You have 8 (NOT 4) equally-likely outcomes. I listed them all. You cannot just distinguish between head #1 and head #2 and ignore all the others: you also need to distinguish between, eg., head#1 and tail#2, head #2 and tail #1, tail #1 and tail #2, etc., etc.
 
  • #13
phinds
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But that is related to what I said in my post. You have 8 (NOT 4) equally-likely outcomes. I listed them all. You cannot just distinguish between head #1 and head #2 and ignore all the others: you also need to distinguish between, eg., head#1 and tail#2, head #2 and tail #1, tail #1 and tail #2, etc., etc.
No, I don't think so. There is a pre-condition that at least one of the coins has a head, so you CANNOT have tail #1 and tail #2. It's ruled out.
 
  • #14
Ray Vickson
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No, I don't think so. There is a pre-condition that at least one of the coins has a head, so you CANNOT have tail #1 and tail #2. It's ruled out.
Yes, I know that. I start with the complete sample space, having 8 equiprobable points. After conditioning, you would be left with 6 (NOT 4) equiprobable points (eliminating (T#1,T#2) and (T#2,T#1)). Among these 6 points, 2 of them correspond to two heads, so we get a conditional probability of 2/6 = 1/3, as before.
 
  • #15
phinds
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OK, I see the error of my ways ... it's 1/3

Thanks
 

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