Step-by-step explanation:
the probability of making one shot is 75% or 0.75 or 3/4.
the probability to make 2 shots is then
3/4 × 3/4 = 9/16 = 0.5625
and so on.
the probability to miss 1 shot is then 25% or 0.25 or 1/4.
the probability to make at least 4 out of 5 shots is the sum of the probability to make all 5 shots plus the probability to make 4 shots.
or the probability to miss 0 plus the probability to miss 1 shot.
anyway, we can go with the positive approach. it seems to be the same complexity as the negative approach.
the probability to make all 5 shots is
(3/4)⁵ = 3⁵/4⁵ = 243 / 1024 = 0.237304688...
the probability to e.g make the first 4 shots and miss the 5th is
(3/4)⁴×1/4 = 3⁴/4⁵ = 81 / 1024 = 0.079101563...
how many possibilities do we have to make 4 out of 5 shots ?
there is no repetition, and the sequence inside the "picked" 4 does not matter, so we need combinations :
C(5,4) = 5! / (4! × (5-4)!) = 5
so, the probability to make exactly 4 out 5 shots is
81/1024 × 5 = 405/1024 = 0.395507813...
and the total probability to make at least 4 shots is
243/1024 + 405/1024 = 648/1024 = 81/128 = 0.6328125
to control,
this plus making exactly 3 plus exactly 2 plus exactly 1 plus none must be the probability 1 (as there is no other possible outcome).
making 3 :
(3/4)³×(1/4)² = 3³/4⁵ = 27/1024
C(5,3) = 5×2 = 10
270/1024
making 2 :
3²/4⁵ = 9/1024
C(5,2) = 5×2 = 10
90/1024
making 1 :
3/1024
C(5,1) = 5
15/1024
making 0 :
1/1024
in sum
(270+90+15+1)/1024 = 376/1024
plus the 648/1024 = 1024/1024 = 1
correct.
