If you toss 10 fair coins in the air, what is the probability that EXACTLY 5 will land on tails?

Question

"If you toss 10 fair coins in the air, what is the probability that EXACTLY 5 will land on tails?"

That would be:

.5^10 = answer, right?
Because, although you need 5 tails...you still need the OTHER 5 to be heads...so therefore you still need all 10 to be a specific result...and a probability of .5 for each(heads or tails)...means .5 x .5 x .5 x .5...etc ... or .5^10, right?

And I don't have a scientific calculator, so could someone give me the decimal and/or percentage answer for this?

Thanks!

Ok, I think I got the answer...

A few comments on how I got to it:

Eric I - You provided the website that gave me the answer. Thank you. But it was not your answer. The website you provided had a thing to input amount of coins, and desired heads...gave a probability of .25...not .50.

Hookodo – I think you had the right idea, but probabilities don’t simplify, I don't think (because of repeatable-possible occurences). 5 heads out of 10 coinflips is not the same as 2 heads out of 4 coinflips (I think. Because the possible UNdesired outcomes – i.e. TTTTTTTTTTH, HTTTTTTTT etc. etc.. increases more than the possible Desired outcome (THTHTHTHTH, or HHHHHTTTTTT….I think....I think, and sorta confirmed by the site)

I think Manjyome had it right...best explanation (although a bit above my comprehension...but I sorta can pick up bits and pieces) and it's confirmed by the site Eric I gave...So Manjyome gets the best answer.

Thanks for all that offered help!

Answer 1

No, because you have to take into account the fact that there is more than one way of getting exactly 5 heads and 5 tails.

For example, when you flip the coin ten times, your outcome might be HHHHH TTTTT, or HTHTH THTH, or HHHTT HHTTT, etc. 0.5^10 is the probability of getting any one specific sequence. You'd have to multiply this by the different numbers of permutations of "5 heads, 5 tails".

If you have a set of n objects were "a" of them are all the same, and "b" of them are all the same, and "c" of them are all the same, etc., then the number of ways you can rearrange them is n! / (a! b! c! ...). For example, the number of ways of arranging the letters in "Mississippi" is 11! / (1! 4! 4! 2!). So the number of ways of getting 5 heads and 5 tails is 10! / (5!5!) = 252. Take this over the total number of outcomes you could have (2^10), and this gives 252 / 2^10 = 252 / 1024 = 63 / 256. This is about 24.6%

I see some people blindly saying "50%", but this is based on the false assumption that "A total of 5 heads and 5 tails" accounts for exactly half of all the possible cases. It actually accounts for much less, as we saw above. It doesn't matter that the coin is fair and the next outcome doesn't depend on the last. We're looking at the whole SET of ten flips.

And I see Supful missed the word "fair" in the first sentence for describing the coins.

Answer 2

Let X be the number of heads seen out of the ten coins tossed. X has the binomial distribution with n = 10 trials and success probability (probability of a head) p = 0.5.

In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[X = x] = 0 for any other value of x.

this is found by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.

the mean of the binomial distribution is n * p
the variance of the binomial distribution is n * p * (1 - p)


for your question.....

P(X = 0) = 0.0009765625 (zero heads)
P(X = 1) = 0.009765625 (exactly 1 head)
P(X = 2) = 0.04394531
P(X = 3) = 0.1171875
P(X = 4) = 0.2050781
P(X = 5) = 0.2460938 (exactly 5 heads)
P(X = 6) = 0.2050781
P(X = 7) = 0.1171875
P(X = 8) = 0.04394531
P(X = 9) = 0.009765625
P(X = 10) = 0.0009765625 (exactly 10 heads)

Answer 3

While there might be some mathematical shortcuts to help derive the numbers, the basic answer is that you have to discover what are all the possibilities, and what are the number of possibilities that will meet the contition.

Let me change to question to demonstrait.

What if you have 4 fair coins and you want to know the odds that exactly two will come up heads.

Well here are all the possibilities

H-H-H-H
H-H-H-T
H-H-T-H
H-H-T-T
H-T-H-H
H-T-H-T
H-T-T-H
H-T-T-T
T-H-H-H
T-H-H-T
T-H-T-H
T-H-T-T
T-T-H-H
T-T-H-T
T-T-T-H
T-T-T-T

You can see that there are 16 possibilities, each on of them just as likely and any other. Of the 16 possibilities, there are 6 ways to have exactly two heads.

So the answer is 6:16, or simplified to the lowest common denominator...
the odds are 3 to 8.

Answer 4

If we assume that an event is impossible, and assign that a value of 0. And we also assume an event will absolutely happen and assign that a value of 1, we have a range to calculate probability. 0 means it cannot happen, and 1 means it will happen.

You want to toss 10 coins at one time and figure the probability of half landing on one side and half landing on the other side. Since the chance of either side ending face up is equal, there are only two possible outcomes, the probability is .5 on the 0 to 1 scale.

The theoretical probability is your event happening is 1 in 2 or .5. However, the laws of probability only apply to large numbers of events. You could produce five tails and five heads on the first toss or not have it happen in 1,000 tosses.

Answer 5

If you toss 10 coins in the air, the total number of possible combinations where 5 come up equals half of the total number of possible combinations coming up.

What you have listed up there is the total chances of one coin coming up heads only or tales only after being tossed 5 times. Check out the web site below for a complete story.

Answer 6

no thats wrong.. since all combinations of heads and tails have equal probablity you need to determine the total number of combinations i.e. 1 tails, 9 heads; 2 tails, 8 heads etc...and then apply that to your combination.. good luck

Answer 7

20

Answer 8

Wouldn't it just be a .5 probability? Becaus no matter how many times you toss it you have a 50/50 chance. The flip before has no influence on the flip after.

Someone flipped a penny over 1000 times and go a .49999 percent occurance of heads, and a .511111 occurance of tails.

Answer 9

well, 0.5^10 is the probability for ONE specific 5,5 combination, like HHHHHTTTTT. but you can always have HTHTHTHTHT or TTTTTHHHHH or many more combination there were. to count how many, the formula is nCr, as i think you must have known.

P = nCr*p^r*q^(n-r)

for n=10, r=5, p=q=0.5

P = 10C5*0.5^5*0.5^5
= 10C5*0.5^10
= 252 / 2^10
= 0.2461
= 24.61%

Answer 10

I would say 1/20.