A student studying for a vocabulary test knows the meanings of 12 words from a list of 26 words. If the test contains 10 words from the study list, what is the probability that at least 8 of the words on the test are words that the student knows? (Round your answer to three decimal places.)

Answer 1

I think this is a binomial distribution question.

prob(knows) = 12/26 = 6/13
prob(not know) = 7/13

prob(students knows at least 8 of 10)
= prob(8of10) + prob(9of10) + prob(10of10)
= C(10,8)(6/13)^8 (7/13)^2 + C(10,9) (6/13)^9 (7/13) + C(10,10) (6/10)^10
= (you do the button pushing)

Answer 2

I CAME UP WITH .03802863003057
however that is wrong.
did exactly as mentions above

(45) (.0020590) (.2899408) === .0268644
+
(10) (.0009503) (.5384615) === .00511699
+
(1) (.0060466) === .0060466

PLEASE HELP

Answer 3

Here you have a small population N=26
and two classes of identical objects, known and not known. Also, the samples are taken without replacement (we don't get to have the same word tested twice in the same test).

Let
K=number of known words = 12
U=number of unknown words = 26-12=14
k=number of known words in the test (sample)
u=number of unknown words in the test = 10-k

Then
P(k)=C(K,k)C(U,u)/C(K+U,k+u)
and
P(8)=C(12,8)C(14,2)/C(26,10)=63/7429
P(9)=C(12,9)C(14,1)/C(26,10)=56/96577
P(10)=C(12,10)C(14,0)/C(26,10)=6/482885
Sum=337/37145=?

Answer 4

you guys are literally no help at all please get smarter.