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

a). 0.032

b). 0.999936

c). 0.00768

d). 0.24576

e). 0.9011

f). 0.3446

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)* (10^(-1))^(6)]`

=
`1-0.000064`

= 0.999936

c). P(4th or 5th) =
`(0.2)^(3)* (0.8)+(0.2)^(4)* (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