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Doubt in probabilities

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Doubt in probabilities [#permalink] New post 09 Feb 2012, 09:27
Hi Bunuel,

I have been reading the MGMAT chapter about probabilities, and they talk about the "OR" questions in which we have to answer what is the probabilty of the Event 1 OR the Event 2.

They present an example in which we have to calculate what is the probability of getting a head at least once in two coin flips. They say that we would be making an error if we said that because we have a 50% chance of getting head on one coin flip, we simply double that chance to 100% for two coin flips. Ok, I agree with that it would be a mistake. However, they say that we have to add the chance in the first flip with the chance in the second flip, and then SUBSTRACT the chance of getting heads on both flips. Algebraicaly:

P(H1 or H2) = P(H1) + P(H2) - P(H1 and H2) = \frac{3}{4}

Here is my doubt: I think that this formula is not to calculate the probability of getting a head al least once in two coin flips. I think that it should be used when we want to calculate the probability of getting ONLY a head in one of the two flips. That's why we substract the event when we get head in the two flips.

If we want to calculate the the probability of getting a head al least once in two coin flips. We have to do this:
H: head; T: Tail
P(H1) = P(HT) = 1/4
P(H2) = P(TH) = 1/4
P(H1 and H2) = P(HH) = 1/4
Adding the three results: 3/4 (the same in the first method)

However, although the result is the same, I think that we are calculating different things. In the first case, we are calculating the chance in getting H1 or H2, but not at the same time, and in the second case we want to calculate the chance of getting at least one head.

What do you think?, or, am I wrong? Please, show me the light! :)
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Re: Doubt in probabilities [#permalink] New post 09 Feb 2012, 09:43
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Personally I'd calculate at least one head with the following approach: {at leas one}=1-{none} --> {at leas one}=1-1/2*1/2=3/4.

Next, MGMAT's using overlapping sets approach for this question: P(H on the first flip)+P(H on the second flip)-P(H on both flips)=1/2+1/2-1/2*1/2. First 1/2 is the probabilities of getting heads on the first flip and second 1/2 is the probabilities of getting heads on the second flip. Now, both include the case of HH so this case is counted twice and that's why the probability of this event is subtracted at the end (first 1/2 is the probability of HT and HH and second 1/2 is the probability of TH and HH).

Hope it's clear.
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Re: Doubt in probabilities [#permalink] New post 10 Feb 2012, 01:21
Expert's post
metallicafan wrote:
Hi Bunuel,

I have been reading the MGMAT chapter about probabilities, and they talk about the "OR" questions in which we have to answer what is the probabilty of the Event 1 OR the Event 2.

They present an example in which we have to calculate what is the probability of getting a head at least once in two coin flips. They say that we would be making an error if we said that because we have a 50% chance of getting head on one coin flip, we simply double that chance to 100% for two coin flips. Ok, I agree with that it would be a mistake. However, they say that we have to add the chance in the first flip with the chance in the second flip, and then SUBSTRACT the chance of getting heads on both flips. Algebraicaly:

P(H1 or H2) = P(H1) + P(H2) - P(H1 and H2) = \frac{3}{4}

Here is my doubt: I think that this formula is not to calculate the probability of getting a head al least once in two coin flips. I think that it should be used when we want to calculate the probability of getting ONLY a head in one of the two flips. That's why we substract the event when we get head in the two flips.

If we want to calculate the the probability of getting a head al least once in two coin flips. We have to do this:
H: head; T: Tail
P(H1) = P(HT) = 1/4
P(H2) = P(TH) = 1/4
P(H1 and H2) = P(HH) = 1/4
Adding the three results: 3/4 (the same in the first method)

However, although the result is the same, I think that we are calculating different things. In the first case, we are calculating the chance in getting H1 or H2, but not at the same time, and in the second case we want to calculate the chance of getting at least one head.

What do you think?, or, am I wrong? Please, show me the light! :)


Even though the question is directed to Bunuel and he has already explained it well, I want to give the diagram here that helps my students understand this concept.
Think of sets.
Attachment:
Ques3.jpg
Ques3.jpg [ 14.39 KiB | Viewed 583 times ]


P(Heads on the first toss) = 1/2 (What does this imply about the second toss? It implies that it doesn't matter what you get on the second toss - heads or tails. Both are acceptable. So it includes the case where you get heads on both the tosses.)
P(Heads on the second toss) = 1/2 (It doesn't matter what you get on the first toss. Both heads and tails are acceptable. So it includes the case where you get heads on both the tosses.)

The case where you get Heads on both the tosses is counted twice. So you have to subtract it once to get the case where you get Heads ATLEAST once.

Think of sets diagram and you will not make a mistake.

Similarly, if you need to find probability of getting heads in ONLY one toss, you can do (P(HT) + P(TH) = 1/4 + 1/4) or if you look at this diagram, you can get it by subtracting the case of both heads twice.
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Re: Doubt in probabilities   [#permalink] 10 Feb 2012, 01:21
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