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A single choice examination has 10 questions, each with five possible answers, only one is correct. Suppose a student randomly guesses at each question.

a.) What is the probability the student answers no questions correct?

b.) What is the probability that the student passes the test (60% or better)

c.) What is the probability that the student's first correct answer is on the seventh question?

My thoughts

a.) (4/5)^10 since the probability of getting each question wrong is the same

b.)

$$P(Y\ge6)=P(Y=6)+ P(Y=4)$$ $$ \left(^{10}_6\right)(.2)^{6}(.8)^{4}+\left(^{10}_{10}\right)(.2)^{10}(.8)^{10-10}$$

$$P(Y\ge6)=.0055$$

c.)(4/5)^6 x (1/5)

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    $\begingroup$ a and c are correct. For b, look up the Bernoulli distribution. If you search the site there are many questions using it. Wikipedia also has a good discussion. $\endgroup$ – Ross Millikan Sep 22 '16 at 21:04
  • $\begingroup$ Is my edit around the right idea? $\endgroup$ – ntgoodatmth Sep 22 '16 at 22:11
  • $\begingroup$ No. The first line says the chance of at least six correct is the sum of the chance of exactly six correct and exactly four correct. The first part of the second line correctly calculates the chance of exactly six correct while the second part is nonsense and does not relate to the chance of exactly four, nor to the chance of more than six. $\endgroup$ – Ross Millikan Sep 22 '16 at 22:15
  • $\begingroup$ I got this from an example in my book. How do I modify it to be correct? $\endgroup$ – ntgoodatmth Sep 22 '16 at 23:04
  • $\begingroup$ John's answer is a good one. His $n=6$ term matches yours $\endgroup$ – Ross Millikan Sep 22 '16 at 23:21
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A straightforward (if long-ish) way to answer the second question is to consider the probability that the person gets exactly $6,7,8,9,$ and $10$ questions correct.

There are ${10 \choose n}$ ways that the student can get $n$ questions correct. The probability that this happens is $(1/5)^n(4/5)^{10-n}$. Then your answer is

$$P_{pass} = \sum_{n=6}^{10}{10 \choose n}\left(\frac15\right)^n\left(\frac45\right)^{10-n}$$

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  • $\begingroup$ is this equivalent to using a binomial distribution formula? $\endgroup$ – ntgoodatmth Sep 22 '16 at 21:48
  • $\begingroup$ As I have showed in my edit $\endgroup$ – ntgoodatmth Sep 22 '16 at 22:05

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