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/