Practice 1 Solutions

1. One thousand people were surveyed online and asked to select their favorite movie genre: action, comedy, animation, or fantasy. As the surveyors reviewed the results, they noticed five people in a row selected “action.”

Independent

2. Sara has a laundry basket full of shirts that need to be hung up in her closet. She grabs a shirt, puts it on a hanger, and hangs it in the closet until the basket is empty.

Dependent

3. Three students in Ainsley’s probability class use a fair, four-section spinner to answer a 50-question multiple-choice test. The scores of each test are compared.

Independent

4. What is the percent chance of flipping a coin and getting tails 5 times in a row?

$\mathrm{P}(\mathrm{T}, \mathrm{T}, \mathrm{T}, \mathrm{T}, \mathrm{T})={\left(\frac{1}{2}\right)}^5=\frac{1}{32}=0.03125\approx 3\%$

3%

5. What is the probability of drawing an ace and then a king from a standard deck of cards when the card is replaced?

$\begin{array}{rcl}\mathrm{P}\left(\mathrm{A}, \mathrm{K}\right)& =& \frac{4}{52}·\frac{4}{52}=\frac{1}{13}·\frac{1}{13}\end{array}$

$\frac{1}{169}$

6. What is the percent chance of rolling the numbers 2, 3, 4, 5 in a row when rolling a six-sided die?

$$\begin{gathered} \mathrm{P}\left(2, 3, 4, 5\right)={\left(\frac{1}{6}\right)}^4=\frac{1}{1296} \\ =0.00077 \end{gathered}$$

0.077%

7. $\mathrm{P}(\mathrm{R}, \mathrm{R}, \mathrm{Y})$

$\begin{array}{rcl}\mathrm{P}\left(\mathrm{R}, \mathrm{R}, \mathrm{Y}\right)& =& \frac{4}{15}·\frac{4}{15}·\frac{2}{15}=\frac{32}{3375}\\ & =& 0.009\overline{481}\end{array}$

0.95%

8. $\mathrm{P}(\mathrm{B}, \mathrm{R}, \mathrm{Y})$

$\begin{array}{rcl}\mathrm{P}\left(\mathrm{B}, \mathrm{R}, \mathrm{Y}\right)& =& \frac{6}{15}·\frac{4}{15}·\frac{2}{15}=\frac{16}{1125}\\ & =& 0.014\overline{2}\end{array}$

1.42%

9. $\mathrm{P}(\mathrm{B}, \mathrm{G}, \mathrm{B})$

$\begin{array}{rcl}\mathrm{P}\left(\mathrm{B}, \mathrm{G}, \mathrm{B}\right)& =& \frac{6}{15}·\frac{3}{15}·\frac{6}{15}=\frac{108}{3375}=\frac{4}{125}\\ & =& 0.032\end{array}$

3.2%

10. $\mathrm{P}(\mathrm{Y}, \mathrm{not} \mathrm{G})$

$\begin{array}{rcl}\mathrm{P}\left(\mathrm{Y}, \mathrm{not} \mathrm{G}\right)& =& \frac{2}{15}·\frac{12}{15}\\ & =& \frac{8}{75}\\ & =& 0.10\overline{6}\end{array}$

10.67%

11. What is the probability that the team will successfully make exactly 16 free throws?

There is a 25.56% chance of successfully making exactly 16 free throws.

12. What is the probability that the team successfully makes at least 16 free throws?

$$\begin{gathered} n=18, r=\left\{16, 17, 18\right\}, p=0.85, q=0.15 \\ \mathrm{P}\left(16\right)=nCr\left(18, 16\right)·{\left(0.85\right)}^{16}·{\left(0.15\right)}^{\left(18–16\right)}=0.2556 \\ \mathrm{P}\left(17\right)=nCr\left(18, 17\right)·{\left(0.85\right)}^{17}·{\left(0.15\right)}^{\left(18–17\right)}=0.1704 \\ \mathrm{P}\left(18\right)=nCr\left(18, 18\right)·{\left(0.85\right)}^{18}·{\left(0.15\right)}^{\left(18–18\right)}=0.0536 \\ \mathrm{P}\left(16\right)+\mathrm{P}\left(17\right)+\mathrm{P}\left(18\right)=0.4796 \end{gathered}$$

47.96%

13. Exactly two items are defective

$$\begin{gathered} n=15, r=2, p=0.03, q=0.97 \\ \mathrm{P}\left(2\right)=nCr\left(15, 2\right)·{\left(0.03\right)}^2·{\left(0.97\right)}^{15–2} \\ =0.0636 \end{gathered}$$

6.36%

15. At most, two items are defective

$$\begin{gathered} n=12, r=\left\{0, 1, 2\right\}, p=0.03, q=0.97 \\ \mathrm{P}\left(2\right)=\mathit{nCr}\left(15, 2\right)·{\left(0.03\right)}^2·{\left(0.97\right)}^{13}=0.0636 \\ \mathrm{P}\left(1\right)=\mathit{nCr}\left(15, 1\right)·\left(0.03\right)1·{\left(0.97\right)}^{14}=0.2938 \\ \mathrm{P}\left(0\right)=\mathit{nCr}\left(15, 0\right)·{\left(0.03\right)}^0·{\left(0.97\right)}^{15}=0.6333 \\ \mathrm{P}\left(2\right)+\mathrm{P}\left(1\right)+\mathrm{P}\left(0\right)=0.9907 \end{gathered}$$

99.07%

16. If there are only 13 items, determine the defect rate if exactly 4 items are defective.

$$\begin{gathered} n=13, r=4, p=0.03, q=0.97 \\ \mathrm{P}\left(4\right)=\mathit{nCr}\left(13, 4\right)·{\left(0.03\right)}^4·{\left(0.97\right)}^{13–4}=0.00044 \end{gathered}$$

0.04%