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.)
you guys are literally no help at all please get smarter.
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=?
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
To find the probability that at least 8 of the words on the test are words that the student knows, we can break it down into two cases: 8 known words and 9 known words.
Case 1: 8 known words
The student needs to choose 8 words they know from the 12 known words and 2 words they don't know from the remaining 14 words in the list. The number of ways to choose these words is given by the combination formula: C(12, 8) * C(14, 2).
Case 2: 9 known words
The student needs to choose 9 words they know from the 12 known words and 1 word they don't know from the remaining 14 words in the list. The number of ways to choose these words is given by the combination formula: C(12, 9) * C(14, 1).
The total number of ways to choose 10 words from the 26-word list is given by C(26, 10).
To compute the probability, we sum up the probabilities of both cases and divide by the total number of ways to choose 10 words:
P(at least 8 known words) = (C(12, 8) * C(14, 2) + C(12, 9) * C(14, 1)) / C(26, 10)
Now let's calculate the probability:
P(at least 8 known words) = [(12! / (8! * (12-8)!)) * (14! / (2! * (14-2)!)) + (12! / (9! * (12-9)!)) * (14! / (1! * (14-1)!))] / (26! / (10! * (26-10)!))
Using a calculator or a programming language that supports factorials and combinations, we can compute the value of this expression. Rounding the answer to three decimal places gives us the desired probability.
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)