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Assuming there is a test a student has to take with multiple choice questions:

  • Each question has 4 choices (a,b,c,d)
  • The student chooses the answer per question randomly
  • There are 100 questions
  • Each question has only 1 correct answer
  • Each question carry out 1 one point if correct

What is the most probable grade for this student?

Option 1: The answer is 25 points. If so, why 25 points (mathematically)?

Option 2: If the answer is 25 points, we ask the question: what is the probability of the same student getting 25 points on this test (a.k.a 1/4 of all question corrects) It will be: (1/4)^25 = 8.88e-16

Which is less probable than getting only one correct, (1/4)^1 = 0.25

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closed as off-topic by Xander Henderson, Claude Leibovici, TheSimpliFire, Clayton, Namaste Aug 10 '18 at 0:22

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    $\begingroup$ What is the distribution of the number of successes in a sequence of (independent) trials with only two outcomes possible? $\endgroup$ – nicomezi Aug 9 '18 at 12:10
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    $\begingroup$ $.25$ is the probability of getting one right if you only try once. The probability of getting exactly one correct if you try $100$ times is $100\times .25^1\times .75^{99}\approx 1.07\times 10^{-11}$ $\endgroup$ – lulu Aug 9 '18 at 12:11
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    $\begingroup$ You may want to read up on the binomial distribution. $\endgroup$ – Xander Henderson Aug 9 '18 at 12:48
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    $\begingroup$ Even though the probability of getting exactly 25 right is small, it is larger than getting any other number right - binomial distribution. $\endgroup$ – herb steinberg Aug 9 '18 at 21:36
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    $\begingroup$ Hint: compute $P(X,\,k+1)/P(X,\,k)$ for $X\sim B(n,\,p)$. $\endgroup$ – J.G. Aug 9 '18 at 22:11
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Comments: I don't know whether you're unsure this is a problem that uses binomial PDFs (or PMFs), or whether you're having trouble computing the numerical values of the formulas. The posted method for the last two computations is not correct.

The only part of the computation that would be difficult on a really cheap calculator would be finding ${100 \choose 25} = \frac{100!}{25! \times 75!}.$ [A statistical calculator should be able to handle such 'binomial coefficients'. Alternatively, perhaps you were supposed to use a normal approximation for that part.]

I have left a few gaps in the computations. I hope you can use this as an outline or for clues.


The number of questions correct by guessing is $X \sim \mathsf{Binom}(n = 100,\, p=1/4).$

Then $\mu_X = E(X) = np = 100(1/4) = 25.$ Also, $\sigma_X^2 = Var(X) = np(1-p) = 75/4 = 18.75$ and $\sigma_X = SD(X) = \sqrt{18.75} \approx 4.33.$


$P(X = 25) = {100 \choose 25}0.25^{25}(1-0.25)^{75}\ \approx 0.0918.$ Computed using the PDF function dbinom in R software:

dbinom(25, 100, .25)
[1] 0.09179969

Computed using R as a calculator:

choose(100,25)*.25^25*.75^75
[1] 0.09179969

Computing this by a normal approximation (with continuity correction and standardization), we have:

$$P(X = 25) = P(24.5 < X < 25.5)\\ =P\left(\frac{24.5 - 25}{4.33} <\frac{X - \mu}{\sigma} < \frac{25.5-25}{4.33}\right)\\ = \cdots \approx P(-0.1155 < Z < 0.1155) \approx 0.0919,$$

where $Z$ has a standard normal distribution. [If you haven't seen this method yet, save this. I predict you will see it soon.]

diff(pnorm(c(24.5,25.5), 25, 4.33))
[1] 0.09193013                 # without standardizing
diff(pnorm(c(-.1155,.1155)))
[1] 0.09195118                 # using standard normal

The figure below shows the exact PDF of $\mathsf{Binom}(100, .25)$ [vertical bars] along with the density curve [blue] of the approximating normal distribution. In answer to the question in the title, it shows that $X=25$ is the most likely outcome. See this Wikipedia section for a more formal discussion of the mode of a binomial distribution. Vertical dashed lines indicate the area under the normal curve that approximates $P(X=25).$

enter image description here


$P(X = 1) = {100 \choose 1}.25^1 .75^{99} = (25).75^{99} \approx 1.07 \times 10^{-11}.$

dbinom(1, 100, .25)
[1] 1.069067e-11
25*.75^99
[1] 1.069067e-11
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