Mastery Check Solutions

1. If all the problems were true/false and Keanu decided to flip a coin to fill in the answers, what is the likelihood that he answers all the problems correctly?

$\mathrm{P}\left(6 \mathrm{correct}\right)={\left(\frac{1}{2}\right)}^6=\frac{1}{64}=0.0156\approx 1.5\%$

2. Explain why the events are independent when guessing.

When guessing, the answer to one problem does not affect the answer to another problem.

3. If all the problems were multiple choice with four answer options each, how likely is it that he would guess the correct answer for all six problems?

$$\begin{gathered} n=6\mathit{,}\mathit{ }r=6\mathit{,}\mathit{ }p=0.25,\mathit{ }q=0.75 \\ \mathit{nCr}\left(6, 6\right)·{\left(0.25\right)}^6·{\left(0.75\right)}^{\left(6–6\right)}=0.0002=0.02\% \end{gathered}$$

4. If all the problems were multiple choice with four answer options each, how likely is Keanu to guess at least four of the six answers correctly?

$$\begin{gathered} \mathrm{Four}: n=6, r=4, p=0.25, q=0.75 \\ \mathrm{F}\mathrm{ive}: n=6, r=5, p=0.25, q=0.75 \\ \mathrm{S}\mathrm{ix}: n=6, r=6, p=0.25, q=0.75 \end{gathered}$$

$$\begin{gathered} \mathit{nCr}\left(6, 4\right)·{\left(0.25\right)}^4·{\left(0.75\right)}^{\left(6–4\right)}=0.0329 \\ \mathit{nCr}\left(6, 5\right)·{\left(0.25\right)}^5·{\left(0.75\right)}^{\left(6–5\right)}=0.0043 \\ \mathit{nCr}\left(6, 6\right)·{\left(0.25\right)}^6·{\left(0.75\right)}^{\left(6–6\right)}=\underline{0.0002} \\ 0.0374=3.74\% \end{gathered}$$

5. Explain to Keanu why guessing is not a good strategy.

Sample: Guessing is not a good strategy because the chance of earning a passing grade on the assignment is very unlikely (less than 4%).