# True false question probability

• Oct 4th 2009, 09:43 PM
rax012
True false question probability
Hello to all,
I have an question that "In 20 questions of true false student require 8 right answer to pass to passout. What is the probability that student will fails?"
• Oct 4th 2009, 10:13 PM
Soroban
Hello, rax012!

Youe grammar is strange . . .

Quote:

In a test of 20 true-false questions, a student needs at least 8 right answers to pass.
What is the probability that student will fail?

I will assume that the student randomly guesses the answers.

To fail, he must have less than 8 right answers.
There are 8 cases to consider . . .

. . $P(\text{0 right}) \:=\:{20\choose0}\left(\frac{1}{2}\right)^0\left(\ frac{1}{2}\right)^{20}$

. . $P(\text{1 right}) \:=\:{20\choose1}\left(\frac{1}{2}\right)^1\left(\ frac{1}{2}\right)^{19}$

. . $P(\text{2 right}) \:=\:{20\choose2}\left(\frac{1}{2}\right)^2\left(\ frac{1}{2}\right)^{18}$

. . $P(\text{3 right}) \:=\:{20\choose2}\left(\frac{1}{2}\right)^3\left(\ frac{1}{2}\right)^{17}$

. . $P(\text{4 right}) \:=\:{20\choose2}\left(\frac{1}{2}\right)^4\left(\ frac{1}{2}\right)^{16}$

. . $P(\text{5 right}) \:=\:{20\choose2}\left(\frac{1}{2}\right)^5\left(\ frac{1}{2}\right)^{15}$

. . $P(\text{6 right}) \:=\:{20\choose2}\left(\frac{1}{2}\right)^6\left(\ frac{1}{2}\right)^{14}$

. . $P(\text{7 right}) \:=\:{20\choose2}\left(\frac{1}{2}\right)^7\left(\ frac{1}{2}\right)^{13}$

$P(\text{7 or less right}) \:=\:\bigg[{20\choose0} + {20\choose1} + {20\choose2} + {20\choose3} + {20\choose4} + {20\choose5} + {20\choose6} + {20\choose7}\bigg]$ $\cdot\left(\frac{1}{2}\right)^{20}$

. . . . . . . . . . . $=\; \bigg[1 + 20 + 190 + 1140 + 4845 + 15,\!504 + 38,\!760\bigg]\cdot \frac{1}{1,\!048,\!576}$

. . . . . . . . . . . $=\;\frac{137,\!980}{1,\!048,\!576} \;=\;\frac{34,\!495}{262,\!144}$