/sci/ - Science & Math

[Anonymous No.968498]

Hi, I need help with one of my probability questions:

A basketball player has an 80% chance of scoring a free throw.
He has 5 shots.
What is the probability that he will score at least 3 times?

What I did was write out all 32 outcomes of INs and OUTs. IIII, OOOOO, IIIIO, etc And count how many time he could get it in 3/5 times, 4/5 times and 5/5 times:
1/32 times for 5/5 times
6/32 times for 4/5 times
10/32 times for 3/5 times
I multiplied those total possibilities by their corresponding probability.
1((5/5)^5) = 0.32768
6(((4/5)^4) * 1/5) = 0.49152
10(3/5 * 3/5 * 3/5 * 2/5 * 2/5) = 0.2048

If the probability of scoring at least 3/5 is the
p(scoring 3/5) + p(scoring 4/5) + p(scoring 5/5)
=1.024

However the answer is supposedly 0.94208

Can anyone here tell me what I did wrong?

[Anonymous No.968511]

>free throw
>five shots

[Anonymous No.968547]

3 / .8

[Anonymous No.968563]

>>968547
wat

[Anonymous No.968571]

It's a probability equation.
You do nCr(number of succes ^ probability of success x number of failures ^ probability of failure).
So for this equation it would be...
5C3 (.8^3)(.2^2)
gg

[Anonymous No.968577]

>>968547
No. I solved it, it turns out I overcounted the 4/5 outcomes, it's 5 not 6. Meaning I can take away 0.08192 from the 1.024 giving me 0.94208

Thanks for anyone who tried /thread

[Anonymous No.968584]

>>968571
You're right about the equation, however the mistake lied in my own ability to count things lol, thanks anyways.

[Anonymous No.968607]

100%

[Anonymous No.968639]

5C3*(.8^3)(.2^2)+5C4(.8^4)(.2)+5C5(.8^5)
10*.512*.04+5*.4096*.2+1*.32768
.2048+.4096+.32768
.94208