Probability problem. Please explain steps. A student studying for a vocabulary test knows the meanings of 10 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.)

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"At least 8 words" can be 8 words, 9 words or 10 words, and these events (possibilities) are mutually exclusive. Therefore if we find the probability `P_8 ` of exactly 8 words, the probability `P_9 ` of exactly 9 words and the probability `P_10 ` of exactly 10 words,...

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Hello!

"At least 8 words" can be 8 words, 9 words or 10 words, and these events (possibilities) are mutually exclusive. Therefore if we find the probability `P_8 ` of exactly 8 words, the probability `P_9 ` of exactly 9 words and the probability `P_10 ` of exactly 10 words, the probability in question will be `P_8 + P_9 + P_10 .`

To find these probabilities, consider the first word in the test. The probability that it is a known word is `10 / 22 . ` The probability that the second word is known provided the first is known is `9 / 21 . ` The probability that the third word is known provided the first and second are known is `8 / 20 , ` and so on.

Thus the probability that the first 8 words are known is
`Pf_8 = 10/22 * 9/21 * 8/20 * 7/19 * 6/18 * 5/17 * 4/16 * 3/15 .`

But 8 words in any order are suitable, so this product must be multiplied by `((10),(8)) = ((10),(2)) = (10*9)/(1*2) = 45, ` so we obtain

`P_8 = 45 * Pf_8 = 45 * 5/11 * 3/7 * 2/5 * 7/19 * 1/3 * 5/17 * 1/4 * 1/5 = `
`= 9 * 1/11 * 1/19 * 5/17 * 1/2 = (9*5)/(11*19*17*2) = 45/7106 approx 0.0063.`

Not so much but `P_9 ` is much less, it is
`P_9 = Pf_8 * 2/14 * ((10),(9)) = Pf_8 * 1/7 * ((10),(1)) = 10/7 * Pf_8 = 2/63 * P_8.`

And `P_10 ` is totally negligible as `P_10 = 1/10 * P_9 * 1/13 = P_8/(63*5*13).`

The sum is about 0.0065. Not a good idea to know so few words...

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