College Algebra, Chapter 10, 10.4, Section 10.4, Problem 16

An archer hits his target $80 \%$ of the time. If he shoots seven arrows, what is the probability of each event?

a.) He never hits the target.

b.) He hits the target each time.

c.) He hits the target more than once.

d.) He hits the target at least five times.

In this case, the probability of success $p=0.80$, then the probability of failure is $q=1-p = 1-0.80=0.20$

a.) If the archer hits the target, then $r=0$, so

$=C(7,0)(0.80)^0 (0.20)^{7-0}$

$=C(7,0) (0.80)^0 (0.20)^7$

$=0.0000128$

b.) If the archer hits the target each time, then $r=7$, so

$=C(7,7)(0.80)^7 (0.20)^{7-7}$

$= C(7,7)(0.80)^7 (0.20)^0$

$=0.2097$

c.) If the archer hits the target more than once, then we get the sum of the probability that he never hits the target and the probability that he hits the target only one $(r-1)$. From part (a), we have

$= 0.000128 + C(7,1)(0.80)^1 (0.20)^{7-1}$

$= 0.000128 + C(7,1) (0.80)(0.20)^6$

$= 0.000128 + 0.0003584$

$= 0.0003712$

Then, by using the complement of probability , the probability that the archer hits the target more than once is $1-0.0003712 = 0.9996288$

d.) The probability that the target is hit at least five times in

$p$(target is hit exactly 5 times) $+ p$(target is hit exactly 6 times) $+ p$(target is hit exactly 7 times)

$=C(7,5)(0.80)^5 (0.20)^{7-5} + C(7,6)(0.80)^6 (0.20)^{7-6} + C(7,7) (0.80)^7 (0.20)^{7-7}$

$= 0.27525 + 0.36700 + 0.20972$

$=0.85197$