Prob of at least 3 heads in 5 coin tosses

[joannabanana]

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]

[GMATGuruNY]

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.

[joannabanana]

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

[GMATGuruNY]

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.

[aimscore]

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

[GMATGuruNY]

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?

[this_time_i_will]

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