If a rifleman hits target 80 percent of the time If he fires four shots what is the probability he will hit the target exactly twice and what is the probability he will hit the target at least twice?

Probabilities are calculated by the binomial distribution. Ans: 0.1536 and 0.9728

Discussion:

Prob of exactly two times = 4!/(2! x 2! ) x (0.8)^2 x (0.2)^2 = 0.1536

Prob of at least twice = hits target 2 times + hits target 3 times + hits target 4 times

Prob of at least twice = 1 - Prob of no hits - Prob of one hit

Prob of one hit = 4!/(3! x 1!) x (0.8)^1 x (0.8)^1 x (0.2)^3 = 0.0256

Prob of no hits= 4!/(0! x 4!) x (0.8)^0 x (0.2)^4 = (0.2)^4 = 0.0016

Prob of 2 or more hits = 1-0.0256-0.0016 = 0.9728

Note: You can calculate these values using Excel, where Prob of 2 hits = binom(2,4,0.8,false)

and Prob of 2 or more hits = 1 - binom(1,4,0.8,false) - binom(0,4,0.8,false)

or:

Prob of 2 > hits = 1- binom(1,4,0.8,true) , as false is requesting the PMF value and true is requesting CDF value. See help in Excel for further explanation on this function.

Also, there might be an issue of independent events in this problem, in that the probability is given as a constant. If a rifleman missed three times, do you think he would learn and might do better than his average on the fourth shot?