Question

what is the probability that a number selected at random from the set {2, 3, 7, 12, 15, 22, 72, 108} will be divisible by both 2 and 3? f. \\(\frac{1}{4}\\) g. \\(\frac{3}{8}\\) h. \\(\frac{3}{5}\\) i. \\(\frac{5}{8}\\) j. \\(\frac{7}{8}\\) in a bag, there are 7 red, 4 white, 33 blue, and 33 yellow cubes. if one of these cubes is selected at random, what is the probability of selecting a cube that is neither blue nor yellow? a. \\(\frac{6}{7}\\) b. \\(\frac{7}{11}\\) c. \\(\frac{1}{3}\\) d. \\(\frac{1}{7}\\)

Explanation:

First Problem: Probability of number divisible by 2 and 3

Step1: Identify total numbers

The set is \(\{2, 3, 7, 12, 15, 22, 72, 108\}\), so total numbers \(n = 8\).

Step2: Find numbers divisible by 2 and 3 (i.e., divisible by 6)

Check each number:

• \(2\): Divisible by 2, not 3.
• \(3\): Divisible by 3, not 2.
• \(7\): Neither.
• \(12\): \(12\div6 = 2\), divisible.
• \(15\): Divisible by 3, not 2.
• \(22\): Divisible by 2, not 3.
• \(72\): \(72\div6 = 12\), divisible.
• \(108\): \(108\div6 = 18\), divisible.

So numbers divisible by 6: \(12, 72, 108\) → count \(m = 3\).

Step3: Calculate probability

Probability \(P=\frac{m}{n}=\frac{3}{8}\).

Step1: Calculate total number of cubes

Red: 7, White: 4, Blue: 33, Yellow: 33. Total \(N = 7 + 4 + 33 + 33 = 77\).

Step2: Find cubes neither blue nor yellow (red + white)

Number of such cubes \(M = 7 + 4 = 11\).

Step3: Calculate probability

Probability \(P=\frac{M}{N}=\frac{11}{77}=\frac{1}{7}\)? Wait, no, wait: Wait, 7+4=11, total is 7+4+33+33=77? Wait, no, 7+4=11, 33+33=66, 11+66=77. Then neither blue nor yellow is red + white = 11. So probability is \(11/77 = 1/7\)? Wait, but let's check again. Wait, 7 red, 4 white: 11. Blue 33, yellow 33: 66. Total 11+66=77. So neither blue nor yellow is 11, so probability is 11/77 = 1/7. Wait, but the options: D is \(1/7\). Wait, let's recheck:

Wait, 7 red, 4 white: 7+4=11. Total cubes: 7+4+33+33=77. So probability is 11/77 = 1/7. So the answer is D. \(\frac{1}{7}\).

Answer:

G. \(\frac{3}{8}\)

Second Problem: Probability of cube neither blue nor yellow