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A multiple choice exam consists of 30 multiple choice questions, with 5 possible answers per question. Assuming that you randomly select an answer for each question, what is the probability that you get exactly 15,16 , or 17 questions correct?

A multiple choice exam consists of 3030 multiple choice questions, with 55 possible answers per question. Assuming that you randomly select an answer for each question, what is the probability that you get exactly 1515,1616 , or 1717 questions correct?

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Q. A multiple choice exam consists of 3030 multiple choice questions, with 55 possible answers per question. Assuming that you randomly select an answer for each question, what is the probability that you get exactly 1515,1616 , or 1717 questions correct?
  1. Understand the problem: Understand the problem and determine the formula to use.\newlineWe need to calculate the probability of getting exactly 1515, 1616, or 1717 questions correct out of 3030 when each question has 55 possible answers. The probability of getting a question right by random guessing is 15\frac{1}{5}, and the probability of getting it wrong is 45\frac{4}{5}. We will use the binomial probability formula:\newlineP(X=k)=(nk)⋅(pk)⋅((1−p)(n−k))P(X = k) = \binom{n}{k} \cdot (p^k) \cdot ((1-p)^{(n-k)})\newlinewhere nn is the number of trials (questions), kk is the number of successes (correct answers), and 161600 is the probability of success on a single trial.
  2. Calculate probability of 1515: Calculate the probability of getting exactly 1515 questions correct.\newlineUsing the binomial formula:\newlineP(X=15)=(3015)⋅(15)15⋅(45)30−15P(X = 15) = \binom{30}{15} \cdot \left(\frac{1}{5}\right)^{15} \cdot \left(\frac{4}{5}\right)^{30-15}\newlineWe calculate (3015)\binom{30}{15} using the combination formula n!/(k!(n−k)!)n! / (k!(n-k)!):\newline(3015)=30!(15!⋅15!)\binom{30}{15} = \frac{30!}{(15! \cdot 15!)}\newlineNow we calculate the probability:\newlineP(X=15)=30!(15!⋅15!)⋅(15)15⋅(45)15P(X = 15) = \frac{30!}{(15! \cdot 15!)} \cdot \left(\frac{1}{5}\right)^{15} \cdot \left(\frac{4}{5}\right)^{15}
  3. Calculate probability of 1616: Calculate the probability of getting exactly 1616 questions correct.\newlineUsing the binomial formula:\newlineP(X=16)=(3016)⋅(15)16⋅(45)30−16P(X = 16) = \binom{30}{16} \cdot \left(\frac{1}{5}\right)^{16} \cdot \left(\frac{4}{5}\right)^{30-16}\newlineWe calculate (3016)\binom{30}{16} using the combination formula n!/(k!(n−k)!)n! / (k!(n-k)!):\newline(3016)=30!16!⋅14!\binom{30}{16} = \frac{30!}{16! \cdot 14!}\newlineNow we calculate the probability:\newlineP(X=16)=30!16!⋅14!⋅(15)16⋅(45)14P(X = 16) = \frac{30!}{16! \cdot 14!} \cdot \left(\frac{1}{5}\right)^{16} \cdot \left(\frac{4}{5}\right)^{14}
  4. Calculate probability of 1717: Calculate the probability of getting exactly 1717 questions correct.\newlineUsing the binomial formula:\newlineP(X=17)=(3017)⋅(15)17⋅(45)30−17P(X = 17) = \binom{30}{17} \cdot \left(\frac{1}{5}\right)^{17} \cdot \left(\frac{4}{5}\right)^{30-17}\newlineWe calculate (3017)\binom{30}{17} using the combination formula n!/(k!(n−k)!)n! / (k!(n-k)!):\newline(3017)=30!17!⋅13!\binom{30}{17} = \frac{30!}{17! \cdot 13!}\newlineNow we calculate the probability:\newlineP(X=17)=30!17!⋅13!⋅(15)17⋅(45)13P(X = 17) = \frac{30!}{17! \cdot 13!} \cdot \left(\frac{1}{5}\right)^{17} \cdot \left(\frac{4}{5}\right)^{13}
  5. Add probabilities: Add the probabilities of getting exactly 1515, 1616, and 1717 questions correct to find the total probability.\newlineP(15,16, or 17 correct)=P(X=15)+P(X=16)+P(X=17)P(15, 16, \text{ or } 17 \text{ correct}) = P(X = 15) + P(X = 16) + P(X = 17)\newlineWe add the probabilities calculated in steps 22, 33, and 44.
  6. Perform calculations: Perform the calculations and simplify the result.\newlineThis step involves actual numerical computation, which typically requires a calculator or computer software, as the factorials and powers involved are too large for manual calculation. However, the process has been described correctly in the previous steps.

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