Question

Need help asap, thanks!!!ヾ(⌐■_■)ノ♪

Answer 1

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.