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Old 10-18-2005, 06:26 PM   #1 (permalink)
Dead Inside
 
Location: East Coast, USA
a simple probability question with real life application

I need to get 24 out of 40 questions in a multiple choice test in order to pass.
I will be able to answer 10 questions, but rest of them I would have to guess blind.
These question have 5 choices so I have 20% chance of getting right.

What is my chance of getting 24 or more right?
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Old 10-18-2005, 07:26 PM   #2 (permalink)
Mjollnir Incarnate
 
Location: Lost in thought
This is what I came up with

You have 100% chance of getting 10 questions right.
That means you only need to correctly answer 14 more.
With a 20% probability of guessing correctly, you would have a probability of (0.2)^14 of passing.

That's a 1.64*10^-8 % chance. Doesn't look good if you're serious (and if I didn't butcher math).
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Old 10-18-2005, 08:07 PM   #3 (permalink)
 
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Location: Waterloo, Ontario
Well, since you already have 10 questions "in the bag," the real question becomes the chance that you get at least 14 questions out of the 30 questions left. If we let <sub>n</sub>C<sub>k</sub> be the choose function, then there are <sub>30</sub>C<sub>14</sub> ways of getting exactly 14 right answers. If there's a 1/5 chance of getting an individual question right, the chances of getting exactly 14 right answers out of 30 are (1/5)<sup>14</sup>&times;(4/5)<sup>30-14</sup>. To address the question of the chances of "or more," we must sum over the chances of the individual numbers. So, the chances of getting 14 or more questions right is <sub>i&isin;[14,30]</sub>&sum;(<sub>30</sub>C<sub>i</sub>&times;(1/5)<sup>i</sup>(4/5)<sup>30-i</sup>).

Of course, this is a pretty wordy formula and my impromptu summation notation probably didn't help readability all that much, but there you have it. You can probably write a script to get a numeric answer, easily enough...
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Old 10-19-2005, 07:40 AM   #4 (permalink)
Mjollnir Incarnate
 
Location: Lost in thought
That's about 9.02*10^-4 or .09%

I should have realized that this called for the binary distribution formula.
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Old 10-19-2005, 09:57 AM   #5 (permalink)
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Location: Moscow on the Ohio
Quote:
Originally Posted by alkaloid
I need to get 24 out of 40 questions in a multiple choice test in order to pass.
I will be able to answer 10 questions, but rest of them I would have to guess blind.
These question have 5 choices so I have 20% chance of getting right.

What is my chance of getting 24 or more right?
Maybe try to improve your odds by assuming that each letter or number a.b.c.d.e or 1,2,3,4,5, will be used approximately an equal number of times as correct answers. After determining which letters or numbers were used the most in the 10 correct answers, select the remainder from those less used. You may be able to tell from this that my test taking skills could use some improvement.
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Old 10-19-2005, 11:41 AM   #6 (permalink)
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Location: Toronto, Ontario, Canada
Quote:
Originally Posted by flstf
Maybe try to improve your odds by assuming that each letter or number a.b.c.d.e or 1,2,3,4,5, will be used approximately an equal number of times as correct answers. After determining which letters or numbers were used the most in the 10 correct answers, select the remainder from those less used. You may be able to tell from this that my test taking skills could use some improvement.
Man, I used to do that. Until I shat my pants at one of my exams. I swear that there were no A's, B's and relatively few c's. The answers were almost all d's and e's.

I left the exam thinking I failed although I studied like mad.

Turned out I got 94%. Go figure.
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Old 10-19-2005, 02:23 PM   #7 (permalink)
Mjollnir Incarnate
 
Location: Lost in thought
My mom took a psych final in college where every question was true/false. Every single answer was false. Very few people passed that one.
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Old 10-25-2005, 06:21 AM   #8 (permalink)
Dead Inside
 
Location: East Coast, USA
Thanks a lot guys. I knew I was in trouble...
I guess .09% is better than 0. I love you all!
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