Question

a coin is weighted so that there is a 57.9% chance of it landing on heads when flipped. the coin is flipped 12 times. find the probability that the number of flips resulting in head is at least 5 and at most 10

Answer 1

Answer)

As there are fixed number of trials and probability of each and every trial is same and independent of each other

Here we need to use the binomial formula

P(r) = ncr*(p^r)*(1-p)^n-r

Ncr = n!/(r!*(n-r)!)

N! = N*n-1*n-2*n-3*n-4*n-5........till 1

For example 5! = 5*4*3*2*1

Special case is 0! = 1

P = probability of single trial = 0.579

N = number of trials = 12

R = desired success = at least 5 and at most 10 = P(5)+P(6)+P(7)+P(8)+P(9)+P(10)

= 12C5*(0.579^5)*(1-0.579)^12-5 +...

= 0.90913956657