Acoin is slightly bent, and as a result the probability of a head is 0.52. suppose that you toss the coin four times. use the binomial formula to find the probability of 3 or more heads

Hi Spor7tdardLamilokab, this particular equation is a binomial probability equation:

P(X) =  * *

Note that it's not n / k, but rather the notation is n choose k. software will not allow me to do this, but imagine the n choose k, without the bar, and that the correct notation.
 
Where equals   , where n! is n * (n-1) * (n-2) 8 ... 1

So for your particular equation:



You would repeat the above equation for 4 as well and add that to the computation for 3, since you want the probability that you'll get 3 or more heads.

Solving this produces the

P(x 3) = 0.34308352

You can also solve this with a TI 83 or 84.

To do this, the steps are 2ND | VARS (or DISTR) | BINOMPDF| TRIALS | PROBABILITY | X VALUE

Doing this for 3 and 4 produces the same result:

binompdf(4, 0.52, 3) + binompdf(4, 0.52, 4) = 0.34308352

Logical equivalent of the give question is its contrapositive