Question

A coin is weighted so that there is a 61.5% chance of it landing on heads when flipped. The coin is flipped 15 times.

Find the probability that the number of flips resulting in "heads" is at least 5 and at most 10.

Answer 1

let us consider p is the probability of  landing on heads = 0.615

n = 15

since this is the case of binomial distribution so

P(5 X10) = P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)

= 0.0189+0.0503+0.1034+0.1651+0.2051+0.1966 = 0.7395 (using binomial calculator)

the probability that the number of flips resulting in "heads" is at least 5 and at most 10. is 0.7395

Also using excel we have

Binomial Probabilities
Data
Sample size	15
Probability of an event of interest	0.615
Statistics
Mean	9.225
Variance	3.5516
Standard deviation	1.8846
Binomial Probabilities Table
	X	P(X)
	5	0.0189
	6	0.0503
	7	0.1034
	8	0.1651
	9	0.2051
	10	0.1966