Question

An Olympic archer is able to hit the bull’s-eye 80% of the time. Assume each shot is independent of the others. If she shoots 6 arrows, what’s the probability of each of the following results?

Answer 1

Answer:

a). 0.032

b). 0.999936

c). 0.00768

d). 0.24576

e). 0.9011

f). 0.3446

Step-by-step explanation:

The given question is incomplete; here is the complete question.

An Olympic archer is able to hit the bull’s-eye 80% of the time. Assume each shot is independent of the others. If she shoots 6 arrows, what’s the probability of each of the following results? a) Her first bull’s-eye comes on the third arrow. b) She misses the bull’s-eye at least once. c) Her first bull’s-eye comes on the fourth or fifth arrow. d) She gets exactly 4 bull’s-eyes. e) She gets at least 4 bull’s-eyes. f) She gets at most 4 bull’s-eyes

a). If archer  shots her first bull's-eye on the third arrow.

Since probability to hit the bull's eye = 80% or 0.80

and probability to miss the bull's eye = 20% or 0.20

So P(miss miss hit) = (0.2)(0.2)(0.8) = 0.032

b). She misses the bull's-eye at least one out of 6 arrows.

So, P(misses at least once) = 1 - P(misses all)

= $1-(0.2)^{6}$

= $1-[(2)^{6}\times (10^{-1})^{6}]$

= $1-0.000064$

= 0.999936

c). P(4th or 5th) = $(0.2)^{3}\times (0.8)+(0.2)^{4}\times (0.8)$

= 0.0064 + 0.00128

= 0.00768

d). For exactly 4 hits,

From the binomial distribution formula,

Binomial probability = $^{n}C_{x}.p^{x}.(1-p)^{n-x}$

P(exactly 4 hits) = $^{6}C_{4}.(0.8)^{4}.(1-0.8)^{2}$

P(exactly 4 hits) = 0.2458

e). She gets at least 4 bull's eyes.

P(x ≥ 4) = $^{6}C_{4}.(0.8)^{4}.(1-0.8)^{2}+^{6}C_{5}.(0.8)^{5}.(1-0.8)^{1}+^{6}C_{6}.(0.8)^{5}.(1-0.8)^{0}$

P(x ≥ 4) = 0.9011

f). She gets at most 4 bull's eyes.

P(at most 4 bull's eyes) =$^{6}C_{0}.(0.8)^{0}.(1-0.8)^{6}+^{6}C_{1}.(0.8)^{1}.(1-0.8)^{5}+^{6}C_{2}.(0.8)^{2}.(1-0.8)^{4}+^{6}C_{3}.(0.8)^{3}.(1-0.8)^{3}+^{6}C_{4}.(0.8)^{4}.(1-0.8)^{2}$= 0.3446