Question

A student studying for a vocabulary test knows the meanings of 14 words from a list of 22 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

The sample space is 22 words and out of it, 14 words are known to have meanings for the student. In the problem, we use 10 C8 + 10 C9 + 10 C10 to associate at least 8 words. The complete formula is (10 C8 * (14/22)^8 * (8/22)^2 + 10 C9  (14/22)^9 * (8/22)^1 + 10 C 10 (14/22)^10 * (8/22)^)0) that is equal to 0.233

Answer 2

At least 8 of the words on the test are words that the student knows means that student knows 8 or 9 or 10 words.

1. Student knows 8 words (2 words are unknown): he knows the meanings of 14 words from a list of 22 words (8 words are unknown), then the number of ways to select 8 known words from 14 and 2 unknown words from 8 is

$C_{14}^8\cdot C_{8}^2=\dfrac{14!}{8!(14-8)!} \cdot \dfrac{8!}{2!(8-2)!}=\dfrac{14!}{8!6!} \cdot \dfrac{8!}{2!6!} =\dfrac{8!\cdot 9\cdot 10 \cdot 11\cdot 12\cdot 13\cdot 14 }{8!\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6}\cdot \dfrac{6!\cdot 7\cdot 8}{6!\cdot 1\cdot2} =3\cdot 11\cdot 13\cdot 7\cdot 7\cdot 4=84084$.

2. Student knows 9 words (1 unknown): he knows the meanings of 14 words from a list of 22 words, then the number of ways to select 9 known words from 14 and 1 from 8 is

$C_{14}^9\cdot C_8^1=\dfrac{14!}{9!(14-9)!}\cdot \dfrac{8!}{1!(8-1)!} =\dfrac{14!}{9!5!}\cdot \dfrac{8!}{1!7!} =\dfrac{9!\cdot 10 \cdot 11\cdot 12\cdot 13\cdot 14 }{9!\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5} \cdot \dfrac{7!\cdot 8}{7!}=11\cdot 13\cdot 14\cdot 8=16016$.

3. Student knows 10 words: he knows the meanings of 14 words from a list of 22 words, then the number of ways to select 10 known words from 14 is

$C_{14}^{10}=\dfrac{14!}{10!(14-10)!} =\dfrac{14!}{10!4!} =\dfrac{10!\cdot 11\cdot 12\cdot 13\cdot 14 }{10!\cdot 1\cdot 2\cdot 3\cdot 4} =11\cdot 13\cdot 7=1001$.

4. He has $C_{22}^{10}=\dfrac{22!}{10!(22-10)!} =\dfrac{22!}{12!10!}=\dfrac{12!\cdot 13\cdot 14\cdot 15\cdot 16\cdot 17\cdot 18\cdot 19\cdot 20\cdot 21\cdot 22}{12!\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot 9\cdot 10} =13\cdot 7\cdot 17\cdot 19\cdot 22= 646646$ ways to select arbitrary 10 words from 22.

5.. The probability that at least 8 of the words on the test are words that the student knows is

$Pr(\text{at least 8 words})=Pr(\text{8 words})+Pr(\text{9 words})+Pr(\text{10 words})=\dfrac{840846+16016+1001}{646646}=0.15634\approx 0.1569$