Zoser wrote:WHat is wrong with this approach?
P(6 at least)= 1-P(not at least 6)
= 1-P(1 and 2 and 3 and 4 and 5 correct answers)
we have 5 correct answers with P= 1/2*1/2*1/2*1/2*1/2= 1/2^5 = 1/32
1-1/32 = 31/32
Why is my approach wrong?
There are a few issues with that solution.
First, you have considered only one outcome in which Brian fails: he gets questions 1, 2, 3, 4, and 5 correct, and the rest incorrect.
What about getting questions 1, 2, 3, 4, and 6 correct, and the rest incorrect?
What about getting questions 3 and 8 correct, and the rest incorrect?
Etc.
Also, your calculation for getting questions 1, 2, 3, 4, and 5 correct (and the rest incorrect) is missing some parts.
P(1, 2, 3, 4, and 5 correct and 6, 7, and 8 incorrect) = P(#1 correct
AND #2 correct
AND #3 correct
AND #4 correct
AND#5 correct
AND #6 INcorrect
AND #7 INcorrect
AND #8 INcorrect)
= P(#1 correct)
x P(#2 correct)
x P(#3 correct)
x P(#4 correct)
xP(#5 correct)
x P(#6 INcorrect)
x P(#7 INcorrect)
x P(#8 INcorrect)
= 1/2
x 1/2
x 1/2
x 1/2
x 1/2
x 1/2
x 1/2
x 1/2
= 1/256
Cheers,
Brent