[math-fun] California Math Exit Exam
Here's another question to add to the California Exit Exam for high school students in mathematics: If there are 80 questions on this test, and each of the 80 questions has 4 choices, what is the probability that someone could pass (>= 55%) [oops, first answer the question of how many questions out of a total of 80 questions 55% is] the test, if for each question he chose one of the 4 answers at random (p=25% for each answer). Alternatively, assuming that the test preparers aren't insane, what is the probability that someone could pass the test by always choosing the "A" answer for each question? Extra credit: using this probability, estimate the number of times one should take the test to raise the probability of passing it at least once to 80%.
I didn't really set up this posting correctly. In California, we now (are supposed to) have an "exit exam" for high school students which sets standards for graduation. http://www.cde.ca.gov/ta/tg/hs/ The math portion of the test is approx. 80 questions, each of which is multiple choice with 4 choices. A radio talk show host here in Los Angeles was carrying on about how easy it should be to pass this examination, since it required only a 55% score to pass. If I've done _my_ math correctly, a completely random answer sheet with 20 questions would pass with a probability of ~ 1/254; with 40 questions would pass with a probability of ~ 1/20560, with 60 questions would pass with a probability of ~ 1/1467711, and with 80 questions would pass with a probability of ~ 10^-8. So, our random student would have to take the test on the order of 10 million times to have a good probability of passing it. So monkeys with typewriters probably aren't going to pass. At 08:42 PM 5/14/2006, Henry Baker wrote:
Here's another question to add to the California Exit Exam for high school students in mathematics:
If there are 80 questions on this test, and each of the 80 questions has 4 choices, what is the probability that someone could pass (>= 55%) [oops, first answer the question of how many questions out of a total of 80 questions 55% is] the test, if for each question he chose one of the 4 answers at random (p=25% for each answer).
Alternatively, assuming that the test preparers aren't insane, what is the probability that someone could pass the test by always choosing the "A" answer for each question?
Extra credit: using this probability, estimate the number of times one should take the test to raise the probability of passing it at least once to 80%.
Date: Mon, 15 May 2006 14:06:57 -0700 From: Henry Baker <hbaker1@pipeline.com>
[...]
The math portion of the test is approx. 80 questions, each of which is multiple choice with 4 choices.
A radio talk show host here in Los Angeles was carrying on about how easy it should be to pass this examination, since it required only a 55% score to pass.
If I've done _my_ math correctly, a completely random answer sheet with 20 questions would pass with a probability of ~ 1/254; with 40 questions would pass with a probability of ~ 1/20560, with 60 questions would pass with a probability of ~ 1/1467711, and with 80 questions would pass with a probability of ~ 10^-8.
So, our random student would have to take the test on the order of 10 million times to have a good probability of passing it.
So monkeys with typewriters probably aren't going to pass.
I get different numbers, though precisely the same conclusion. (If I've got _my_ math right -- that's what peer review is for, no? :-) Let's assume the number of questions answered correctly is binomially distributed: 80 Bernouilli trials, 25% success rate on each trial, pass the exam with 44 questions right. Then the answer to your question is the CUMULATIVE binomial probability, i.e., the probability of getting q or fewer questions right, in whatever order. It's a one-liner in R:
pbinom(q = 0:80, size = 80, prob = 0.25, lower.tail = FALSE) [1] 1.000000e+00 1.000000e+00 1.000000e+00 9.999997e-01 9.999977e-01 [6] 9.999877e-01 9.999460e-01 9.997991e-01 9.993523e-01 9.981607e-01 [11] 9.953407e-01 9.893589e-01 9.778938e-01 9.579033e-01 9.260137e-01 [16] 8.792424e-01 8.159061e-01 7.364254e-01 6.436978e-01 5.428363e-01 [21] 4.402937e-01 3.426341e-01 2.553323e-01 1.819482e-01 1.238525e-01 [26] 8.047434e-02 4.988718e-02 2.949574e-02 1.662971e-02 8.939674e-03 [31] 4.581986e-03 2.239142e-03 1.043316e-03 4.635213e-04 1.963610e-04 [36] 7.931939e-05 3.055204e-05 1.122084e-05 3.929250e-06 1.311755e-06 [41] 4.174442e-07 1.266114e-07 3.659175e-08 1.007433e-08 2.641417e-09 [46] 6.593068e-10 1.565976e-10 3.537701e-11 7.597276e-12 1.549988e-12 [51] 3.002147e-13 5.516119e-14 9.606372e-15 1.584139e-15 2.471005e-16 [56] 3.641559e-17 5.063669e-18 6.633998e-19 8.175502e-20 9.460183e-21 [61] 1.025786e-21 1.039938e-22 9.832241e-24 8.644741e-25 7.045311e-26 [66] 5.302663e-27 3.670229e-28 2.324706e-29 1.339769e-30 6.978136e-32 [71] 3.258192e-33 1.350390e-34 4.907634e-36 1.539764e-37 4.086262e-39 [76] 8.919250e-41 1.537460e-42 1.962434e-44 1.648989e-46 6.842278e-49 [81] 0.000000e+00
These are the probabilities that q > n (strictly greater than). Check: Pr(q > 0) = 1 (yeah, we oughta be able to do that...) Pr(q > 79) = 0.25^80 = 6.8e-49 Median is 20: qbinom(p = 0.5, size = 80, prob = 0.25, lower.tail = FALSE) == 20 So you get 0.55 * 80 = 44 answers correct with probability around 2.6e-09. Now, how many times do you need to take the test in order to pass it 1ce on the last iteration? Let's assume that's governed by the geometric distribution: k-1 failures, followed by exactly 1 pass on the last try. (Sort of like fertility treatment: you treat until the first pregnancy, then STOP!) The probability we pass after k iterations or fewer is then the cumuative geometric probability:
pgeom(q = 1:50, prob = 2.641417e-09) [1] 5.282834e-09 7.924251e-09 1.056567e-08 1.320708e-08 1.584850e-08 [6] 1.848992e-08 2.113134e-08 2.377275e-08 2.641417e-08 2.905559e-08 [11] 3.169700e-08 3.433842e-08 3.697984e-08 3.962125e-08 4.226267e-08 [16] 4.490409e-08 4.754550e-08 5.018692e-08 5.282834e-08 5.546976e-08 [21] 5.811117e-08 6.075259e-08 6.339401e-08 6.603542e-08 6.867684e-08 [26] 7.131826e-08 7.395967e-08 7.660109e-08 7.924251e-08 8.188392e-08 [31] 8.452534e-08 8.716676e-08 8.980817e-08 9.244959e-08 9.509101e-08 [36] 9.773242e-08 1.003738e-07 1.030153e-07 1.056567e-07 1.082981e-07 [41] 1.109395e-07 1.135809e-07 1.162223e-07 1.188638e-07 1.215052e-07 [46] 1.241466e-07 1.267880e-07 1.294294e-07 1.320708e-07 1.347123e-07
Uh-oh... doesn't look good even after 50 tries! We can compute the median number of tries to pass at random, using the geometric quantile function:
qgeom(p = 0.5, prob = 2.641417e-09) [1] 262414900
So, no monkeys pass unless you have immortal monkeys typing at lightspeed. (Did I really just say that?) If the above passes the peer review of Math-Fun, perhaps you could forward a suitably anonymized version to the radio talk show host. (Or you could ask yourself why you're listening to talk radio? :-) -- Steve Rowley <sgr@alum.mit.edu> http://alum.mit.edu/www/sgr/
Steve, et al: My reasoning is as follows: If "r" is the probability of getting an answer right, and "w" (= 1-r) is the probability of getting an answer wrong, then prob(passing) = sum(r^(n-i)*w^i*binomial(n,n-i),i,0,.45*n), n=# of questions This gives us r^n+binomial(n,n-1)*r^(n-1)*w+binomial(n,n-2)*r^(n-2)*w^2+... +binomial(n,n-j)*r^(n-j)*w^j where j is the maximum number one can get wrong = .45*n. Since r=1/4, w=1-r=3/4, so this summation simplifies to sum((1/4)^(n-i)*(3/4)^i*binomial(n,n-i),i,0,.45*n) = sum((1/4)^n*1^(n-i)*3^i*binomial(n,n-i),i,0,.45*n) = (1/4)^n*sum(3^i*binomial(n,n-i),i,0,.45*n) = 2^(-2*n)*sum(3^i*binomial(n,n-i),i,0,.45*n) = 2^-160*sum(3^i*binomial(80,80-i),i,0,36) = 7361825474021217224187948500896604842715 ------------------------------------------------ 730750818665451459101842416358141509827966271488 ~ 1.0074330792356688E-8, according to Maxima Is this correct? At 07:59 AM 5/16/2006, Steve Rowley wrote:
Date: Mon, 15 May 2006 14:06:57 -0700 From: Henry Baker <hbaker1@pipeline.com>
[...]
The math portion of the test is approx. 80 questions, each of which is multiple choice with 4 choices.
A radio talk show host here in Los Angeles was carrying on about how easy it should be to pass this examination, since it required only a 55% score to pass.
If I've done _my_ math correctly, a completely random answer sheet with 20 questions would pass with a probability of ~ 1/254; with 40 questions would pass with a probability of ~ 1/20560, with 60 questions would pass with a probability of ~ 1/1467711, and with 80 questions would pass with a probability of ~ 10^-8.
So, our random student would have to take the test on the order of 10 million times to have a good probability of passing it.
So monkeys with typewriters probably aren't going to pass.
I get different numbers, though precisely the same conclusion. (If I've got _my_ math right -- that's what peer review is for, no? :-)
Let's assume the number of questions answered correctly is binomially distributed: 80 Bernouilli trials, 25% success rate on each trial, pass the exam with 44 questions right. Then the answer to your question is the CUMULATIVE binomial probability, i.e., the probability of getting q or fewer questions right, in whatever order.
It's a one-liner in R:
pbinom(q = 0:80, size = 80, prob = 0.25, lower.tail = FALSE) [1] 1.000000e+00 1.000000e+00 1.000000e+00 9.999997e-01 9.999977e-01 [6] 9.999877e-01 9.999460e-01 9.997991e-01 9.993523e-01 9.981607e-01 [11] 9.953407e-01 9.893589e-01 9.778938e-01 9.579033e-01 9.260137e-01 [16] 8.792424e-01 8.159061e-01 7.364254e-01 6.436978e-01 5.428363e-01 [21] 4.402937e-01 3.426341e-01 2.553323e-01 1.819482e-01 1.238525e-01 [26] 8.047434e-02 4.988718e-02 2.949574e-02 1.662971e-02 8.939674e-03 [31] 4.581986e-03 2.239142e-03 1.043316e-03 4.635213e-04 1.963610e-04 [36] 7.931939e-05 3.055204e-05 1.122084e-05 3.929250e-06 1.311755e-06 [41] 4.174442e-07 1.266114e-07 3.659175e-08 1.007433e-08 2.641417e-09 [46] 6.593068e-10 1.565976e-10 3.537701e-11 7.597276e-12 1.549988e-12 [51] 3.002147e-13 5.516119e-14 9.606372e-15 1.584139e-15 2.471005e-16 [56] 3.641559e-17 5.063669e-18 6.633998e-19 8.175502e-20 9.460183e-21 [61] 1.025786e-21 1.039938e-22 9.832241e-24 8.644741e-25 7.045311e-26 [66] 5.302663e-27 3.670229e-28 2.324706e-29 1.339769e-30 6.978136e-32 [71] 3.258192e-33 1.350390e-34 4.907634e-36 1.539764e-37 4.086262e-39 [76] 8.919250e-41 1.537460e-42 1.962434e-44 1.648989e-46 6.842278e-49 [81] 0.000000e+00
These are the probabilities that q > n (strictly greater than).
Check: Pr(q > 0) = 1 (yeah, we oughta be able to do that...) Pr(q > 79) = 0.25^80 = 6.8e-49 Median is 20: qbinom(p = 0.5, size = 80, prob = 0.25, lower.tail = FALSE) == 20
So you get 0.55 * 80 = 44 answers correct with probability around 2.6e-09.
Now, how many times do you need to take the test in order to pass it 1ce on the last iteration? Let's assume that's governed by the geometric distribution: k-1 failures, followed by exactly 1 pass on the last try. (Sort of like fertility treatment: you treat until the first pregnancy, then STOP!) The probability we pass after k iterations or fewer is then the cumuative geometric probability:
pgeom(q = 1:50, prob = 2.641417e-09) [1] 5.282834e-09 7.924251e-09 1.056567e-08 1.320708e-08 1.584850e-08 [6] 1.848992e-08 2.113134e-08 2.377275e-08 2.641417e-08 2.905559e-08 [11] 3.169700e-08 3.433842e-08 3.697984e-08 3.962125e-08 4.226267e-08 [16] 4.490409e-08 4.754550e-08 5.018692e-08 5.282834e-08 5.546976e-08 [21] 5.811117e-08 6.075259e-08 6.339401e-08 6.603542e-08 6.867684e-08 [26] 7.131826e-08 7.395967e-08 7.660109e-08 7.924251e-08 8.188392e-08 [31] 8.452534e-08 8.716676e-08 8.980817e-08 9.244959e-08 9.509101e-08 [36] 9.773242e-08 1.003738e-07 1.030153e-07 1.056567e-07 1.082981e-07 [41] 1.109395e-07 1.135809e-07 1.162223e-07 1.188638e-07 1.215052e-07 [46] 1.241466e-07 1.267880e-07 1.294294e-07 1.320708e-07 1.347123e-07
Uh-oh... doesn't look good even after 50 tries! We can compute the median number of tries to pass at random, using the geometric quantile function:
qgeom(p = 0.5, prob = 2.641417e-09) [1] 262414900
So, no monkeys pass unless you have immortal monkeys typing at lightspeed. (Did I really just say that?)
If the above passes the peer review of Math-Fun, perhaps you could forward a suitably anonymized version to the radio talk show host. (Or you could ask yourself why you're listening to talk radio? :-) -- Steve Rowley <sgr@alum.mit.edu> http://alum.mit.edu/www/sgr/
Date: Tue, 16 May 2006 12:10:03 -0700 From: Henry Baker <hbaker1@pipeline.com>
My reasoning is as follows: [... Henry reasons ...] ~ 1.0074330792356688E-8, according to Maxima
Yes, I agree. (In my previous answer I made a fencepost error looking at the next entry in a table.) You did your calculation from first principles; I did mine by recognizing the distribution. Since it seems I do statistics for a living nowadays, that's the quicker route to an answer. BTW, I wrote up the problem and my proposed answers here: http://alum.mit.edu/www/sgr/california-math-exit-exam.html Feel free to correct any mistakes! -- Steve Rowley <sgr@alum.mit.edu> http://alum.mit.edu/www/sgr/ Skype: sgr000
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