What is the probability of getting only 2 heads out of 14 coin flips?

Question

I can't believe I'm asking this. It should be an easy one.

OK at least there's one smart guy amidst the great unwashed...

And one guy who is a smartass as well as being smart. Thanks shlomo. That is the answer I'm looking for. You are indeed wise.

Just goes to show girls (including Mr. Mom) don't get math...

It also goes to show my sister cannot pick football games against the spread...

Funny stuff sunshine jelly. You are obviously a skilled tosser.

Answer 1

0.00555


(14*13/2)/2^14

Answer 2

Hi,
The responder who used Pascal's triangle gave you a correct method, but I'll just add a few comments and other ways to find the answer.
This is basically a binomial distribution problem, and the easy way is to use the binompdf( function on a graphing calculator. I'll cover that later, but first let's look at some other ways to do it.
First of all, in elementary statistics books, there's usually a table in the back of the book that allows you to look up the answer if you have n, p, and r. Here n = number of trials, p = probability of one trial, and r = number of successes. In your case n = 14, p=0.5, and r = 2. So, you can look it up.:-)
The next method is to use the binomial formula. It's this:
P(r) = nCr*P^r*q^(n-r)
Now, this may look quite formidable, but actually it's pretty easy. The only term that you don't know is q and that's just 1-p = 0.5
So, let's plug our numbers in this formula and see what it looks like.
P(2) = 14C2(0.5)²(0.5)^12
Now, you can do combinations on just about any scientific calculator and all graphing calculators.
P(2) = 91(.25)(2.4414E-4)
Plug that in your calculator and you get:
P(2) = 0.00555
Now, the really easy way is to do it on a graphing calculator. I'll give the method for the TI-83 Plus or TI-84. They're both the same.

Calculator:
a) Press DISTR (that's the VARS button), press ALPHA; then A. That'll past binompdf( to the home screen.
b) Enter the number of trials, 14, the probability for one trial, .5, and the number of successes, 2. So that you have this: binompdf(14, .5, 2
You can either close the parentheses on not, it doesn't matter.
c) Press ENTER and the answer .00555 will be displayed.

Hope this helps.
FE

Answer 3

It isn't. Let's not assume the coin is fair. The probability of heads is p and tails is 1-p.

The answer is 14-choose-2 x (p)^2 x (1-p)^12

If the coin is fair (p = 1/2), this simplifies to...

14-choose-2 x (p)^14

What is 14-choose-2? It is the second entry in the fourteenth row of Pascal's triangle. It is 14*13/2 or 91

91 * (1/2)^14 = 91/16384

Answer 4

Let X be the number of heads.

X has the binomial distribution with n = 14 trials and success probability of 0.5 (assuming a fair coin)

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

P(X = 2) = 14! / (2! * (14-2)! ) * 0.5^2 * (1-0.5)^(14-2)
P(X =2) = 0.005554199

Answer 5

That is impossible to tell since those random patterns apply only to large numbers of trials and 14 is not a large number.

If you had said "2 million out of 14 million" the answer would be an obvious 1 out of 7 because that has a large number of attempts in it.

Answer 6

let us assume it's a fair coin.
Probability of getting a head = 0.5 or 1/2
probability of a tail (no head) is 0.5
This is a Binomial distribution with n=14 (number of flips) p=0.5 (probability of a head)
and q=(1-p)=0.5 (probability of a tail)
p(x=2) = C(14,2) x (0.5)^2 (1-0.5)^12
C(14,2)=14!/2! 12! = 0.0055

Answer 7

That's a hard one. How many flips in the air does the coin make before the end result? did you start with heads up or down? Dumb question deserves a dumb answer.

Answer 8

It depends on whether or not the coin was heads up when you flipped it.

i.e.
place the coin on the thumb, heads up... flip it... you will see it will land on tails.

place the coin on the thumb, tails up... flip it.. you will see it will land on heads

Answer 9

err.. 2 over 14 times 100%? logically it all depends on luck isn't it

Answer 10

14C2*(1/2^14)

=91/16384