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

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=(14!)/(8!(14-8)!) \cdot (8!)/(2!(8-2)!)=(14!)/(8!6!) \cdot (8!)/(2!6!) =(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 (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=(14!)/(9!(14-9)!)\cdot (8!)/(1!(8-1)!) =(14!)/(9!5!)\cdot (8!)/(1!7!) =(9!\cdot 10 \cdot 11\cdot 12\cdot 13\cdot 14 )/(9!\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5) \cdot (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)=(14!)/(10!(14-10)!) =(14!)/(10!4!) =(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)=(22!)/(10!(22-10)!) =(22!)/(12!10!)=(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})=(840846+16016+1001)/(646646)=0.15634\approx 0.1569`

Answer 2

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