Question

mary shots free throws 70% on average. She is fouled shooting a three point shot and therefore gets 3 free throws. What is the probability she:

*makes first misses seconds and third shots
*Makes at least 1 shot

Answer 1

h -- hit
m -- miss

she has the following cases:

hhh
hhm
hmh
mhh
hmm *
mhm
mmh
mmm


"makes first misses seconds and third shots " --- I see one case, marked with *

so prob(hmm) = 1/8

"Makes at least 1 shot"
all except mmm
so prob(of that event) = 1 - 1/8 = 7/8

Answer 2

ignore my previous answers, I totally forgot about the actual probabilities.

so for
prob(makes first misses seconds and third shots)
= (7/10)(3/10)(3/10)
= 63/1000 or .063

prob(missing all 3) = (3/10)^3 = 27/1000
so prob(of making at least one shot)
= 1 - 27/1000
= 973/1000
= .973