A coin 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) = \(( \frac{n}{k})\) \(p^{k}\) \((1 - p)^{n-k}\)
Note that it's not n / k, but rather the notation is n choose k. Brainly software will not allow me to do this, but imagine the n choose k, without the bar, and that the correct notation.
Where \((\frac{n}{k})\) equals \(\frac{n!}{k!(n-k)!}\) , where n! is n (n-1) (n-2) 8 ... 1
So for your particular equation:
\(P(x \geq 3) = (\frac{5}{3}) * 0.52^{3} * (1 - .52)^{5-3}\)
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 answer:
P(x\(\geq\) 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