# Binomial Questions

The Spring test will contain $20$ multiple choice questions, each with five responses, and only one response is correct for each question. No marks are deducted for incorrect answers.

1. If a student randomly guesses the answer to every question, calculate the probability that he fails the exam (i.e. scores below $40\%$).
2. If a student is confident on $5$ questions and guesses all of the remaining questions, what is the probability that he passes?
3. If a student is confident on $18$ questions and guesses the remaining questions, what is the probability that he scores $100\%$?

I just want to double check to see if my thinking is correct:

1. $20×40\% = 8$.

$$P(\text{score below}~40\%) = ~^{20}C_0\cdot\left(\frac{1}{5}\right)^0\cdot\left(\frac{4}{5}\right)^{20} + ~^{20}C_1 \cdots + ~^{20}C_8\cdot\left(\frac{1}{5}\right)^8\cdot\left(\frac{4}{5}\right)^{12}$$

1. $20-5 = 15$.

$$P(\text{passing the exam})= ~^{15}C_1\cdot\left(\frac{1}{5}\right)^1\cdot\left(\frac{4}{5}\right)^{14} + ~^{15}C_2 \cdots + ~^{15}C_3\cdot\left(\frac{1}{5}\right)^3\left(\frac{4}{5}\right)^{12}.$$

1. $20 - 18 = 2$.

$$P(\text{scores}~100\%)= ~^2C_2\cdot\left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right)^0.$$

Many thanks!

• @Alex Francisco thank you for the edit – user538621 Mar 8 '18 at 5:44

$(i)$

If the student gets $8$ questions correct, then they score a $40$%, but the question asks for the probability that the student scores $below$ $40$%. Thus we have

$$\sum_{k=0}^7 {20 \choose k}\left(\frac{1}{5}\right)^k\left(\frac{4}{5}\right)^{20-k}\approx0.968$$

$(ii)$

You tried to calculate the probability that the student $fails$, but the question asks for the probability of $passing$.

The student must get $3$ or more of the remaining $15$ questions correct to pass, which is the compliment of getting $0,1,$ or $2$ correct. Thus we have

$$1-\sum_{k=0}^2 {15 \choose k}\left(\frac{1}{5}\right)^k\left(\frac{4}{5}\right)^{15-k}\approx0.602$$

Alternatively, you could have done

$$\sum_{k=3}^{15} {15 \choose k}\left(\frac{1}{5}\right)^k\left(\frac{4}{5}\right)^{15-k}\approx0.602$$

$(iii)$

The student must get the two remaining questions correct with probability $$0.2^2=0.04$$

which you correctly obtained.

• Ahhh that makes sense I forgot that getting the 8th question means that he passes the exam so 8th question is not included! Thank you! – user538621 Mar 8 '18 at 5:44
• Simple mistake! – Remy Mar 8 '18 at 5:44
• So many typos made. I think I have fixed them all. – Remy Mar 8 '18 at 5:48