Question

1. fiona has a bag of marbles that are all the same size and shape. the bag contains 3 red marbles, 4 blue marbles, 3 green marbles, 4 yellow marbles, and 1 white marble. what is the probability of randomly choosing a white marble on the first pick, not replacing it, and then randomly choosing a green marble on the second pick? a \\(\frac{4}{29}\\) b \\(\frac{1}{15}\\) c \\(\frac{1}{75}\\) d \\(\frac{1}{70}\\)

Explanation:

Step1: Find total number of marbles

First, we calculate the total number of marbles. The bag has 3 red, 4 blue, 3 green, 4 yellow, and 1 white marble. So total marbles \( n = 3 + 4 + 3 + 4 + 1 = 15 \).

Step2: Probability of white first (no replacement)

The probability of choosing a white marble first is \( P(\text{white first})=\frac{\text{number of white marbles}}{\text{total number of marbles}}=\frac{1}{15} \). After removing one white marble, the total number of marbles becomes \( 15 - 1 = 14 \).

Step3: Probability of green second

The number of green marbles is 3. So the probability of choosing a green marble second (after removing one white marble) is \( P(\text{green second})=\frac{\text{number of green marbles}}{\text{new total number of marbles}}=\frac{3}{14} \).

Step4: Probability of both events

Since these are dependent events, we multiply the probabilities: \( P = P(\text{white first}) \times P(\text{green second})=\frac{1}{15} \times \frac{3}{14}=\frac{3}{210}=\frac{1}{70} \). Wait, but let's re - check the total number of marbles again. Wait, the problem statement says "Fiona has a bag of marbles that are all the same size and shape. The bag contains 3 red marbles, 4 blue marbles, 3 green marbles, 4 yellow marbles, and 1 white marble." So total marbles: \( 3 + 4+3 + 4+1=15 \). Probability of white first: \( \frac{1}{15} \). Then, after taking out white, total marbles are 14, green marbles are 3. So \( \frac{1}{15}\times\frac{3}{14}=\frac{3}{210}=\frac{1}{70} \). But wait, maybe I misread the red marbles? Wait, no, the problem says "3 red marbles, 4 blue marbles, 3 green marbles, 4 yellow marbles, and 1 white marble". So total is 3 + 4+3 + 4+1 = 15. Then the probability of white first is \( \frac{1}{15} \), then green second is \( \frac{3}{14} \), multiply them: \( \frac{1\times3}{15\times14}=\frac{3}{210}=\frac{1}{70} \). But let's check the options. Option D is \( \frac{1}{70} \). Wait, but maybe I made a mistake in total marbles? Wait, 3 (red) + 4 (blue) + 3 (green) + 4 (yellow) + 1 (white) = 15. Yes. So the probability is \( \frac{1}{15}\times\frac{3}{14}=\frac{1}{70} \).

Answer:

D. \(\frac{1}{70}\)