Suppose that an unfair coin comes up heads 569 of the time T

Suppose that an unfair coin comes up heads 56.9% of the time. The coin is flipped a total of 27 times.

a) What is the probability that you get at most 25 heads?

b) What is the probability that you get exactly 12 tails?

Solution

We will use bonomial theoram to solve this

B(x; n, P) = _nC_x * P^x * (1 – P)^{n – x}

Where:
B = binomial probability
x = total number of “successes” (pass or fail, heads or tails etc.)
P = probability of a success on an individual trial
n = number of trials

a) The probability that getting at most 25 heads

So if we subtract probability of exactly 27 and 26 heads for total probability i.e. 1 we will get probability of at most 25 heads

P(h) = 0.569

P(t) = 1- 0.569 = 0.431

Probability of 27 heads B(27;27,0.569) = 27C₂₇ * 0.569²⁷ * 0.431^{0 =} 1* 0.00000024436*1 =0.00000024436

Probability of 26 heads B(26;27,0.569) = 27C₂₆ * 0.569²⁶ * 0.431^{1 =} 27* 0.000000429456*0.431 =0.0000049975

Probability that getting at most 25 heads = 1 - 0.00000024436 - 0.0000049975

= 1 - 0.00000524

= 0.99999476

b) The probability that you get exactly 12 tails

Probability of 12 tails B(12;27,0.431) = 27C₁₂ * 0.569^27-12 * 0.431¹²

=17383860*0.00021217*0.00004109

= 0.15155