# Edward has to take a five question multiple-choice quiz and

Edward has to take a five question multiple-choice quiz and his socialolgy class. each question asked four choices for answers of which only one is correct. assuming that edwards guesses on all five questions. what is the probability that he will answer a) all five questions correctly b) exactly 2 questions correctly c) at least two questions correctly

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habbocowji
Step 1
Given,
Total number of questions $$\displaystyle={5}$$
Probability of success $$\displaystyle={\frac{{{1}}}{{{4}}}}$$ ( since each question has only 1 correct choice out of 4 choices)
We use binomial distribution here.
Step 2
(a) The probability that he will answer all five questions correctly:
$$\displaystyle{P}{\left({X}={5}\right)}={5}{C}_{{{5}}}{\left({\frac{{{1}}}{{{4}}}}\right)}^{{{5}}}{\left({1}-{\frac{{{1}}}{{{4}}}}\right)}^{{{5}-{5}}}$$
$$\displaystyle={0.000977}$$
(b) The probability that he will answer exactly 2 questions correctly:
$$\displaystyle{P}{\left({X}={2}\right)}=^{{{5}}}{C}_{{{2}}}{\left({\frac{{{1}}}{{{4}}}}\right)}^{{{2}}}{\left({1}-{\frac{{{1}}}{{{4}}}}\right)}^{{{5}-{2}}}$$
$$\displaystyle={0.26367}$$
(c) The probability that he will answer at least 2 questions correctly:
$$\displaystyle{P}{\left({X}\geq{2}\right)}={1}-{P}{\left({X}{ < }{2}\right)}$$
$$\displaystyle={1}-{P}{\left({X}={0}\right)}+{P}{\left({X}={1}\right)}$$
$$\displaystyle={1}-{\left[{0.2373}+{0.3955}\right]}$$
$$\displaystyle={1}-{0.6328}$$
$$\displaystyle={0.3672}$$
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autormtak0w

Step 1
a) The probability that Edward will answer all the six questions correctly is,
$$P\text{(Six correct answers)}=((6),(6))(\frac{1}{4})^{6}(\frac{3}{4})^{0}$$
$$\displaystyle\approx{0.0002}$$

Step 2
Thus, the probability that Edward will answer all the six questions correctly is 0.0002.
b) The probability that Edward will answer exactly two questions correctly is,
$$\displaystyle{P}\text{(Two correct answers)}={\left(\begin{array}{c} {6}\\{2}\end{array}\right)}{\left({\frac{{{1}}}{{{4}}}}\right)}^{{{2}}}{\left({\frac{{{3}}}{{{4}}}}\right)}^{{{4}}}$$
$$\displaystyle\approx{0.2966}$$
Step 3
Thus, the probability that Edward will answer exactly two questions correctly is 0.2966.
c) The probability that Edward will answer at least two questions correctly is,
$$\displaystyle{P}\text{(At least two correct answers)}={P}{\left({X}\geq{2}\right)}$$
$$\displaystyle={1}-{P}{\left({X}{ < }{2}\right)}$$
$$\displaystyle={1}-{P}{\left({X}\le{1}\right)}$$
$$\displaystyle={1}-{\left[{P}{\left({X}={0}\right)}+{P}{\left({X}={1}\right)}\right]}$$
$$\displaystyle={1}-{\left({0.1780}+{0.3560}\right)}$$
$$\displaystyle={0.4660}$$
Thus, the probability that Edward will answer at least two questions correctly is 0.4660.

nick1337
I think the answer is c, at least three questions are correct, because he cannot answer everything correctly, and he cannot get the perfect number, 3.