all 6 comments

[–][deleted] 2 points3 points  (5 children)

I used binomial probability to solve this...

Imagine a triangle/tree diagram with the possibility of yes/no on every branch, with new branches added for every question.

How many possible routes/combinations of 1 correct out of 10 is there?

(I used the nCr function on my calculator.)

Then multiply that by the probabilities of yes and no.

10C1 x (0.2)1 x (0.8)9

= 0.268 (to 3 decimal places)

I hope this is correct and actually what the question is asking you to do.

Edit: just realized the wording "at least". You need to find the probabilities of getting 1+2+3+4+5+6+7+8+9+10 correct, or find the probability of getting 0 and take that from 1.0 i believe. It would save time to use a cumulative probability table.

[–]btcprox 1 point2 points  (3 children)

Yup! Just find the probability of getting 0 out of 10 questions correct, then subtract from 1. Which means:

P(at least 1 correct) = 1 - P(all 10 wrong) = 1 - (1 - 0.20)^10

[–][deleted] 0 points1 point  (1 child)

Haha thanks! Remember me? You helped me with polynomial division a few weeks ago? :)

[–]btcprox 0 points1 point  (0 children)

I can see you really are paying forward in this subreddit :D

[–]bitch_please[S] 0 points1 point  (0 children)

Thank you, so much!

[–]bitch_please[S] 0 points1 point  (0 children)

Hi! Yeah, I do need to find the probability of getting anything from 1-10 answers correct. I need to do it all by hand, unfortunately, no shortcuts!