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]

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] When we flip a coin 5 times, we get either: 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.

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

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 Here's the formulaic (longer) approach. 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.

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

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 When a coin is flipped 5 times, there are only 6 possible combinations of head and tails: 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?

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 Atleast 3 heads = heads 3 times OR heads 4 times OR heads 5 times 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