Shelia Lawrence

2021-12-19

In addition to his social studies class, Edward must complete a five question multiple-choice exam. There were four possible answers for each question, but only one of them was right. assuming Edwards guesses on each of the five inquiries. What is the likelihood that he will respond correctly to all five questions, exactly two questions, or at least two questions?

habbocowji

Step 1
Given,
Total number of questions $=5$
Probability of success $=\frac{1}{4}$ ( 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:
$P\left(X=5\right)=5{C}_{5}{\left(\frac{1}{4}\right)}^{5}{\left(1-\frac{1}{4}\right)}^{5-5}$
$=0.000977$
(b) The probability that he will answer exactly 2 questions correctly:
$P\left(X=2\right){=}^{5}{C}_{2}{\left(\frac{1}{4}\right)}^{2}{\left(1-\frac{1}{4}\right)}^{5-2}$
$=0.26367$
(c) The probability that he will answer at least 2 questions correctly:
$P\left(X\ge 2\right)=1-P\left(X<2\right)$
$=1-P\left(X=0\right)+P\left(X=1\right)$
$=1-\left[0.2373+0.3955\right]$
$=1-0.6328$
$=0.3672$

autormtak0w

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

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

nick1337