A multiple-choice exam consists of 10 questions, each with 5 possible answers. All questions carry equal weight and 60% is the lowest passing score.If a student will be picking answers at random what's the probability of getting a passing score?
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Anonymous
ANSWERS: 2
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Do your own homework. OK, the chance of getting any particular question right is 20%. In order to get a score of 60% or more, the student has to get 6, 7, 8, 9, or 10 right. Do your own homework. The number of possibilities of getting 6 right is 1^6 * 5 * 5 * 5 * 5, or 625/(5^10)or 0.000064, or .0064%, or 64 chances in 10,000, or 4 in 625. Do your own homework. For 7 right, it's 1^7 * 5 * 5 * 5, or 125/9765625, or .0000128. Add the decimals together for 6, 7, 8, 9 and 10 right, and you will have the probability. I bet it will come out about .013%. Or not - I guess you will have to do the work. :o)
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if you need 60%, you need atleast 6 questions right. so, what are the ways to get 10, 9, 8, 7, or 6 questions answered right? 10) there is a 1/5 chance to answer any one questions right, and you need all of them answered right, so (1/5)^10 chance of getting a all 10 right 9) (1/5)^9 seems to be the intuitive probability for this, however there are 10 ways to get this result (10 different choices for which answer will be wrong) so the correct probability here is 10*(1/5)^9 8) there are 10 slots, and you choose 2 (so it's a combination, and not a permutation) 10C2 = 10!/((10-2)!*2!) and we multiply that number by (1/5)^8 I bet you can figure out 6 and 7 now, multiply: P(10)*P(9)...P(6) and you'll have your answer
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