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Question

Mathematics Question on Probability

On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

Answer

p=13\frac{1}{3} and q=1-p=1−13=231-\frac{1}{3}=\frac{2}{3}

n=5,r=4, 5 and P(X=r)=nCrprqn−r^nC_rp^rq^{n-r}

P (four or more success)=P(X=4)+P(X=5)

=3C4(13)4(23)1+3C3(13)5=5∗2∗13+(13)4=11243^3C_4\bigg(\frac{1}{3}\bigg)^4\bigg(\frac{2}{3}\bigg)^1+^3C_3\bigg(\frac{1}{3}\bigg)^5=\frac{5*2*1}{3}+\bigg(\frac{1}{3}\bigg)^4=\frac{11}{243}