Susan Nall

2021-12-17

Beth is taking an eleven​-question ​multiple-choice test for which each question has three answer​ choices, only one of which is correct. Beth decides on answers by rolling a fair die and marking the first answer choice if the die shows 1 or​ 2, the second if the die shows 3 or​ 4, and the third if the die shows 5 or 6. Find the probability of the stated event.
exactly
four

Jenny Bolton

Expert

Step 1
Introduction:
In a fair die, there are 6 possible outcomes, 1, 2, ..., 6, all of which are equally likely, that is, with probability $\frac{1}{6}$.
Step 2
Calculation:
On rolling the fair die once, the probability of getting 1 or 2 is $\frac{1}{3}\left[=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}\right]$.
Similarly, the probability of getting 3 or 4 is 1/3, and the probability of getting 5 or 6 is 1/3.
Thus, in case of each of the 11 questions, the probability of selecting each option is 1/3.
As the outcomes on the roll of a fair die are independent from one roll to another, the choices made in the different questions are also independent of one another.
Considering each attempted question as a trial, there are $n=11$ independent trials.
Considering it to be a success if the correct option is chosen, the probability of success in each trial is $p=\frac{1}{3}$, as each question has only one correct option.
Here, X can be considered as the number of successes, that is, number of questions correctly answered. Thus, X has a binomial distribution with parameters, $n=11,p=\frac{1}{3}$.
Since p denotes the probability of success, the probability of failure is, $q=1-p=\frac{2}{3}$.
If $X\sim Binomial\left(n,p\right)$, then the probability mass function of X is:
$f\left(x\right)=\left\{\begin{array}{ll}\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}{q}^{n-x};x=0,1,...,n;& 0\end{array}$ Here, n is the number of independent trials, p is the probability of success in each trial, and q is the probability of failure.
The probability mass function of X here is:
$f\left(x\right)=\left\{\begin{array}{ll}\left(\begin{array}{c}11\\ x\end{array}\right)\left(\frac{1}{3}{\right)}^{x}\left(\frac{2}{3}{\right)}^{11-x};& x=0,1,2,...,11;\\ 0;& otherwise\end{array}$
The probability of exactly 4 correct answers, that is, exactly 4 successes is calculated below:
$\left(X=4\right)=f\left(4\right)$
$=\left(\begin{array}{c}11\\ x\end{array}\right)\left(\frac{1}{3}{\right)}^{4}\left(\frac{2}{3}{\right)}^{11-4}$
$=\left(330\right)\cdot \left(\frac{1}{81}\right)\cdot \left(\frac{128}{2187}\right)$
$\approx 0.238446$.
Thus, the probability of exactly 4 correct answers is 0.238446.

Debbie Moore

Expert

Step 1
Binomial Problem with $n=10$ and $p\left(c\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}rect\right)=\frac{1}{3}$
Find the probability of

b) fewer than three correct answers

nick1337

Expert

Step 1
Hence, the probability of getting the correct answer is
$p=\frac{1}{3}=0.33333333$, as there are 3 choices (the die part simply shows she is guessing).
Note that the probability of x successes out of n trials is

where
$n=\text{number of trials}=8$
$p=\text{the probability of a success}=0.333333333$
$x=\text{the number of successes}=6$
Thus, the probability is
P(6)=0.017070569

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