Question

use this for #11 - 14: a bag contains 25 red markers, 18 blue markers, and 29 yellow markers. a m 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

Explanation:

Step1: Calculate the total number of markers

The total number of markers is \(25 + 18+29=72\).

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

Since the marker is replaced, the two events are independent.
The probability of choosing a red marker \(P(\text{red})=\frac{25}{72}\), and the probability of choosing a blue marker \(P(\text{blue})=\frac{18}{72}\).

Step3: Calculate \(P(\text{red then blue})\)

For independent events \(A\) and \(B\), \(P(A\text{ and }B)=P(A)\times P(B)\).
So \(P(\text{red then blue})=\frac{25}{72}\times\frac{18}{72}=\frac{25\times18}{72\times72}=\frac{450}{5184}=\frac{25}{288}\)

Step4: Calculate \(P(\text{two blues})\)

The probability of choosing a blue marker in the first draw \(P(\text{blue}_1)=\frac{18}{72}\), and since the marker is replaced, the probability of choosing a blue marker in the second draw \(P(\text{blue}_2)=\frac{18}{72}\)
\(P(\text{two blues})=\frac{18}{72}\times\frac{18}{72}=\frac{324}{5184}=\frac{1}{16}\)

Answer:

1. D. \(\frac{25}{288}\)
2. A. \(\frac{1}{16}\)