(THIS IS A REVIEW QUESTION SO HELP IS ALLOWED)
A bag contains 5 blue marbles, 6 red marbles, 7 green marbles, and 4 white marbles.
If two marbles are selected randomly, find the following probabilities as reduced
fractions:
If the first is replaced, Red A/121 A
Without replacement, Blue and then a Green
If the first is replaced, Black
Without replacement, White A

Ask by O'Quinn Rodriquez. in Canada

Mar 24,2025

Question

(THIS IS A REVIEW QUESTION SO HELP IS ALLOWED)
A bag contains 5 blue marbles, 6 red marbles, 7 green marbles, and 4 white marbles.
If two marbles are selected randomly, find the following probabilities as reduced
fractions:
If the first is replaced, Red A/121 A
Without replacement, Blue and then a Green
If the first is replaced, Black
Without replacement, White A

Ask by O'Quinn Rodriquez. in Canada

Mar 24,2025

UpStudy AI · Accepted Answer

$$
\textbf{Step 1. Total marbles: } 5 + 6 + 7 + 4 = 22.
$$

$$
\textbf{(a) With replacement, } P(2 \text{ Red}):
$$

The probability of selecting a red marble on one draw is

$$
\frac{6}{22} = \frac{3}{11}.
$$

Since the marble is replaced, the probability of drawing two reds is

$$
\left(\frac{3}{11}\right)^2 = \frac{9}{121}.
$$

$$
\textbf{(b) Without replacement, } P(\text{Blue then Green}):
$$

- The probability that the first marble is blue:

$$
\frac{5}{22}.
$$

- After drawing a blue marble, there are $$22-1=21$$ marbles left. The probability that the next marble is green is

$$
\frac{7}{21} = \frac{1}{3}.
$$

Therefore, the probability of drawing a blue then a green is

$$
\frac{5}{22} \times \frac{1}{3} = \frac{5}{66}.
$$

$$
\textbf{(c) With replacement, } P(2 \text{ Black}):
$$

Since there are no black marbles in the bag, the probability is

$$
0.
$$

$$
\textbf{(d) Without replacement, } P(2 \text{ White}):
$$

- The probability that the first marble is white:

$$
\frac{4}{22}.
$$

- After drawing a white marble, the number of white marbles left is 3 and the total marbles left is $$21$$. Thus, the probability that the second marble is white is

$$
\frac{3}{21} = \frac{1}{7}.
$$

Thus, the probability of drawing two white marbles without replacement is

$$
\frac{4}{22} \times \frac{1}{7} = \frac{4}{154} = \frac{2}{77}.
$$

$$
\textbf{Final Answers:}
$$
- $$ P(2 \text{ Red}) = \frac{9}{121} $$
- $$ P(\text{Blue then Green}) = \frac{5}{66} $$
- $$ P(2 \text{ Black}) = 0 $$
- $$ P(2 \text{ White}) = \frac{2}{77} $$
$$ \textbf{Probabilities:} $$ - $$ P(2 \text{ Red}) = \frac{9}{121} $$ - $$ P(\text{Blue then Green}) = \frac{5}{66} $$ - $$ P(2 \text{ Black}) = 0 $$ - $$ P(2 \text{ White}) = \frac{2}{77} $$

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