Question

use this for #11 - 14: a bag contains 25 red markers, 18 blue markers, and 29 yellow markers. a marker is selected, replaced, then another marker is chosen.

1. p(red then blue)

a. 25284 b. 30201 c. 1975184 d. 25288

1. p(two blues)

a. 116 b. 14 c. 110 d. 18

1. mason chooses a card from hearts. probability its a face card?

a. 14 b. 352 c. 313 d. 713

1. lauren chooses a date in september (1 - 30). probability she picks a two - digit number or an even number?

a. 1130 b. 56 c. 65 d. 0

Explanation:

Step1: Calculate total number of markers

Total markers \(=25 + 18+29=72\)

Step2: Calculate \(P(\text{red then blue})\)

Since the marker is replaced, the two events are independent.
\(P(\text{red})=\frac{25}{72}\), \(P(\text{blue})=\frac{18}{72}\)
\(P(\text{red then blue})=P(\text{red})\times P(\text{blue})=\frac{25}{72}\times\frac{18}{72}\)

$$
LATEXBLOCK0
$$

Step3: Calculate \(P(\text{two blues})\)

\(P(\text{blue})=\frac{18}{72}=\frac{1}{4}\)
\(P(\text{two blues})=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}\)

Step4: Calculate probability of face - card from hearts

In a standard deck, there are 13 hearts. Face - cards (Jack, Queen, King) in hearts: 3
\(P(\text{face card from hearts})=\frac{3}{13}\)

Step5: Calculate probability of two - digit number or even number in September dates (1 - 30)

Two - digit numbers: \(10 - 30\) (21 numbers). Even numbers: \(2,4,\cdots,30\) (15 numbers). Two - digit even numbers: \(10,12,\cdots,30\) (11 numbers)
Using the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
\(P(\text{two - digit})=\frac{21}{30}\), \(P(\text{even})=\frac{15}{30}\), \(P(\text{two - digit and even})=\frac{11}{30}\)
\(P(\text{two - digit or even})=\frac{21 + 15-11}{30}=\frac{25}{30}=\frac{5}{6}\)

Answer:

1. D. \(25/288\)
2. A. \(1/16\)
3. C. \(3/13\)
4. B. \(5/6\)