If the coin is tossed 5 times, what is the probability that at least 3 out of 5 times it will show heads?
[spoiler]ans = 1/2[/spoiler]
Prob of at least 3 heads in 5 coin tosses
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- joannabanana
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When we flip a coin 5 times, we get either:joannabanana wrote:If the coin is tossed 5 times, what is the probability that at least 3 out of 5 times it will show heads?
[spoiler]ans = 1/2[/spoiler]
Heads 3 times, tails 2 times
Heads 4 times, tails 1 time
Heads 5 times, tails 0 times
OR
Tails 3 times, heads 2 times
Tails 4 times, heads 1 time
Tails 5 times, heads 0 times
Thus P(heads at least 3 times) = P(tails at least 3 times) = 1/2.
Last edited by GMATGuruNY on Thu Oct 07, 2010 12:41 pm, edited 2 times in total.
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- joannabanana
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Oh wow, that's really clever!
I was looking for a more formula-type answer, but this is way better. I doubt I'd be able to come up with something like this myself though, especially while writing the test. But I guess that's what these forums are for
I was looking for a more formula-type answer, but this is way better. I doubt I'd be able to come up with something like this myself though, especially while writing the test. But I guess that's what these forums are for
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Here's the formulaic (longer) approach.joannabanana wrote:Oh wow, that's really clever!
I was looking for a more formula-type answer, but this is way better. I doubt I'd be able to come up with something like this myself though, especially while writing the test. But I guess that's what these forums are for
P(HHHTT) = (1/2)^5 = 1/32.
We need to multiply by the number of ways to arrange HHHTT = 5!/3!*2! = 10.
10*1/32 = 10/32.
P(HHHHT) = (1/2)^5 = 1/32.
Since T could appear in any of the 5 positions, we multiply by 5.
5*1/32 = 5/32.
P(HHHHH) = (1/2)^5 = 1/32.
Since all of the above are good outcomes, we add the fractions:
10/32 + 5/32 + 1/32 = 16/32 = 1/2.
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When a coin is flipped 5 times, there are only 6 possible combinations of head and tails:aimscore wrote:Hi Gmatguruny,
Pardon my ignorance but can u please explain how u reached this statement ?
Thus P(heads at least 3 times) = P(tails at least 3 times) = 1/2.
ie how did you get the value 1/2
Thanks
In the following 3 combinations, we get heads at least 3 times:
Heads 3 times, tails 2 times
Heads 4 times, tails 1 time
Heads 5 times, tails 0 times
In the following 3 combinations, we get tails at least 3 times.
Tails 3 times, heads 2 times
Tails 4 times, heads 1 time
Tails 5 times, heads 0 times
The probability of getting the first set of outcomes is the same as the probability of getting the second set of outcomes.
Thus, P(heads at least 3 times) = 1/2 and P(tails at least 3 times) = 1/2.
Does this help?
Last edited by GMATGuruNY on Thu Oct 07, 2010 7:21 pm, edited 1 time in total.
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Atleast 3 heads = heads 3 times OR heads 4 times OR heads 5 timesjoannabanana wrote:Oh wow, that's really clever!
I was looking for a more formula-type answer, but this is way better. I doubt I'd be able to come up with something like this myself though, especially while writing the test. But I guess that's what these forums are for
the total number of outcomes in 5 tosses = 2^5 = 32
heads 3 times = 5c3/32 = 10/32
heads 4 times = 5c4/32 = 5/32
heads 5 times = 5c5/32 = 1/32
adding we get 16/32 = 1/2
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