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). 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)

=

=

=

= 0.999936

c). P(4th or 5th) =

= 0.0064 + 0.00128

= 0.00768

d). For exactly 4 hits,

From the binomial distribution formula,

Binomial probability =

P(exactly 4 hits) =

P(exactly 4 hits) = 0.2458

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

P(x ≥ 4) =

P(x ≥ 4) = 0.9011

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

P(at most 4 bull's eyes) == 0.3446

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation:

false

step-by-step explanation: