Question

A survey of high school juniors found that 82% of students plan on attending college. If you pick three students at random, what is the probability that at least two plan on attending college? Round to the nearest percent.
I continue to get 36%, but the answer is 91%. How?

Answer 1

I used this formula:
\[P(x)= nC _{x}p ^{x}q ^{n-x}\]
with the first 'n' as a subscript. This is the only formula included in the lesson, so I assumed it was the correct one to use on this problem.

Answer 2

okk i finally figured it out lol

Answer 3

this is how u do it!!
lets look at the solution that u dont want
the 3 persons u pick
___ , ____, _____
no no no = 0.18^3
no no yes = 0.18 * 0.18 * 0.82
no yes no = 0.18*0.82*0.18
yes no no = 0.82*0.18*0.18

so in other words all the ones u dont need = 0.18^3+3 * (0.18^2*0.82)
now the probability of the success = just 1 - 0.18^3+3 * (0.18^2*0.82)
= 91.44%

Answer 4

the other way to do it wud be to look at all the success so like
----, -----, -----
yes yes yes = 0.82^3
yes no yes =0.82^2*0.18
yes yes no =0.82^2*0.18
no yes yes =0.82^2*0.18

add all that up = 0.914464
or 91.44% hope this helps there is a formula u can form out of this too like
P of success ^ random choices + (P of success ^ random choices - 1) * number of choices )

Answer 5

Thank you so much!

Answer 6

np :)

Answer 7

also if it may help you in the future the way i arrived at the fact that those only 8 were the possible outcomes was that
-----, -----,----- 2*2*2 = 8 like a coin flip except the sides have a different probability of landing. I know it doesnt sound very useful when its only for 3 people and 2 choices, but if these numbers pile up it would be a different story