A blindfolded marksman finds that on average he hits the target 70% of the time. If he fires 6 shots, determine the probability of the following 3 situations(round answers to the nearest thousandths):

Question

more than two hits

two hits or less

exactly two hits

Answer 1

You can solve this using the binomial theorem.

P(0 hits) = (.3)^6 =.000729

P(1 hit) = 6(.7)(.3)^5 = .010206

P(2 hits) = 6C2 (.7)^2 (.3)^4 = .059535

You can take it from there

Answer 2

Can use the binomial expansion of ( x+y)^6 with x representing hitting the target (p=.7) and y represents not hitting the target (p=0.3). You can get the coefficients of each of these terms using Pascals triangle or combinatorics

The term x^2 y^4 represents him hitting the target exactly twice. Coefficient of this term is 15 so this probability is 15 ( 0.7)^2 (0.3)^4= .0595

Two hits or less would be 2, 1 or no hits. We already have the probability of 2 hits

Probability of 1 hit is the x y^5 term. Coefficient of this term is 6 so probability of exactly 1 hit is

(6) (0.7) ( 0.3^5) = .010206

Probability of no hits is (0.3)^6 = .000729

If we add these three together we will get the probability of 2 hits or less

.0595 = .010206 + .000729= .072289 or 7.229%

Probability of more than 2 hits would be 100% - 7.229% =92.77%