Stephen answers 4 mc questions by wild guessing . If each question contains 4 choices in which only one is correct , find the probability that Stephen gets
(i)only three questions correct
(ii)one answer wrong
p.s expain with answers
f5 maths
2007-02-21 1:17 am
回答 (2)
2007-02-21 1:31 am
✔ 最佳答案
i)Probability of guessing right = 1/4 per mc question
Probability of guessing wrong = 3/4 per mc question
As the probability is only one is in wrong, we could use:
(1/4)(1/4)(1/4)(3/4) = 3/256
As there will have 4 difference combination.
So:
( 3/256 ) * 4
= 3/64
ii)
Answer is same as question i) Since -
Probability of guessing right = 1/4 per mc question
Probability of guessing wrong = 3/4 per mc question
2007-02-21 1:25 am
(i) Pr(correct ans | one question) = 1/4
Pr(3 questions correct) = Pr (1st, 2nd, 3rd correct and 4th wrong) + Pr (1st, 3rd, 4th correct and 2nd wrong) +Pr (1st, 2nd, 4th correct and 3rd wrong) +Pr (2nd, 3rd, 4th correct and 1st wrong)
= 1/4*1/4*1/4*3/4+1/4*1/4*1/4*3/4+1/4*1/4*1/4*3/4+1/4*1/4*1/4*3/4=3/64
(ii) Pr(1 ans wrong) = Pr (3 questions correct) = 3/64
Pr(3 questions correct) = Pr (1st, 2nd, 3rd correct and 4th wrong) + Pr (1st, 3rd, 4th correct and 2nd wrong) +Pr (1st, 2nd, 4th correct and 3rd wrong) +Pr (2nd, 3rd, 4th correct and 1st wrong)
= 1/4*1/4*1/4*3/4+1/4*1/4*1/4*3/4+1/4*1/4*1/4*3/4+1/4*1/4*1/4*3/4=3/64
(ii) Pr(1 ans wrong) = Pr (3 questions correct) = 3/64
參考: myself
收錄日期: 2021-04-12 23:23:11
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