An Olympic archer is able to hit the bull’s-eye 15% of the time. Assume each shot is independent of the others. If she shoots 7 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) 0.108

b) 1.00

c) 0.170

d) 0.0109

e) 0.0121

f) 0.999

Step-by-step explanation:

Success: p = 0.15

Failure: q = 1 - p = 0.85

a) q²p = 0.85² × 0.15

= 0.108375 = 0.108 (3 sf)

b) 1 - P(all hits)

= 1 - p⁷ = 1 - 0.15⁷

= 0.9999982914= 1.00 (3 sf)

c) q³p + q⁴p = 0.1704196875

= 0.170 (3 sf)

d) 7C4×p⁴×q³ = 35 × 0.15⁴ × 0.85³

= 0.01088152734 = 0.0109 (3 sf)

e) P(4,5,6 or 7)

= 7C4×p⁴×q³ + 7C5×p⁵×q² +

7C6×p⁶×q + 7C7×p⁷×q⁰

= 0.01210317188 = 0.0121 (3 sf)

f) 1 - P(more than 4)

= 1-(0.01210317188-0.01088152734)

= 1 - 0.001221644531

= 0.9987783555 = 0.999 (3 sf)

i only know the sequence of arithmetic it adds the first is 5+4 then it goes up as it adds to then it is 9+5 then 14+6 then to 20+7 and so on. 5,9,14,20,27,35.

step-by-step explanation: