A bag contains $4$ blue and $8$ green marbles. Three marbles are selected at random from the bag. $\newline$ a. If the first marble is replaced before the second marble is drawn what is $\mathrm{P}$ (blue second I green first)? Use the conditional probability formula! $\newline$ b. If the first marble is NOT replaced what is P(blue second / green first)? Use conditional probability formula!

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Algebra 2

Find probabilities using the normal distribution I

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Q. A bag contains

4

blue and

8

green marbles. Three marbles are selected at random from the bag.

\newline

a. If the first marble is replaced before the second marble is drawn what is

\mathrm{P}

(blue second I green first)? Use the conditional probability formula!

\newline

b. If the first marble is NOT replaced what is P(blue second / green first)? Use conditional probability formula!

Calculate Probability with Replacement: a. Calculate the probability of drawing a blue marble second given that a green marble was drawn first with replacement. $\newline$ Since the first marble is replaced, the probability of drawing a blue marble remains the same for the second draw.
Calculate Probability without Replacement: The total number of marbles is $4$ blue + $8$ green = $12$ marbles. $\newline$ The probability of drawing a blue marble is the number of blue marbles divided by the total number of marbles, which is $\frac{4}{12}$ or $\frac{1}{3}$ .
Calculate Probability without Replacement: The total number of marbles is $4$ blue + $8$ green = $12$ marbles. The probability of drawing a blue marble is the number of blue marbles divided by the total number of marbles, which is $\frac{4}{12}$ or $\frac{1}{3}$ .Using the conditional probability formula $P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$ , we need to find $P(\text{blue second} | \text{green first})$ . Since the first marble is replaced, the events are independent, and $P(A \text{ and } B) = P(A) \times P(B)$ . However, since we are given that a green marble is drawn first, $P(B)$ is certain ( $P(B) = 1$ ), and thus $8$ $0$ .
Calculate Probability without Replacement: The total number of marbles is $4$ blue + $8$ green = $12$ marbles. The probability of drawing a blue marble is the number of blue marbles divided by the total number of marbles, which is $\frac{4}{12}$ or $\frac{1}{3}$ . Using the conditional probability formula $P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$ , we need to find $P(\text{blue second} | \text{green first})$ . Since the first marble is replaced, the events are independent, and $P(A \text{ and } B) = P(A) \times P(B)$ . However, since we are given that a green marble is drawn first, $P(B)$ is certain ( $P(B) = 1$ ), and thus $8$ $0$ . Therefore, $P(\text{blue second} | \text{green first})$ with replacement is simply the probability of drawing a blue marble, which is $\frac{1}{3}$ or approximately $8$ $3$ .
Calculate Probability without Replacement: The total number of marbles is $4$ blue + $8$ green = $12$ marbles. $\newline$ The probability of drawing a blue marble is the number of blue marbles divided by the total number of marbles, which is $\frac{4}{12}$ or $\frac{1}{3}$ .Using the conditional probability formula $P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$ , we need to find $P(\text{blue second} | \text{green first})$ . $\newline$ Since the first marble is replaced, the events are independent, and $P(A \text{ and } B) = P(A) \times P(B)$ . $\newline$ However, since we are given that a green marble is drawn first, $P(B)$ is certain ( $P(B) = 1$ ), and thus $8$ $0$ .Therefore, $P(\text{blue second} | \text{green first})$ with replacement is simply the probability of drawing a blue marble, which is $\frac{1}{3}$ or approximately $8$ $3$ .b. Calculate the probability of drawing a blue marble second given that a green marble was drawn first without replacement. $\newline$ Since the first marble is not replaced, the total number of marbles decreases by one, and the number of green marbles decreases by one.
Calculate Probability without Replacement: The total number of marbles is $4$ blue + $8$ green = $12$ marbles. $\newline$ The probability of drawing a blue marble is the number of blue marbles divided by the total number of marbles, which is $\frac{4}{12}$ or $\frac{1}{3}$ .Using the conditional probability formula $P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$ , we need to find $P(\text{blue second} | \text{green first})$ . $\newline$ Since the first marble is replaced, the events are independent, and $P(A \text{ and } B) = P(A) \times P(B)$ . $\newline$ However, since we are given that a green marble is drawn first, $P(B)$ is certain ( $P(B) = 1$ ), and thus $8$ $0$ .Therefore, $P(\text{blue second} | \text{green first})$ with replacement is simply the probability of drawing a blue marble, which is $\frac{1}{3}$ or approximately $8$ $3$ . $\newline$ b. Calculate the probability of drawing a blue marble second given that a green marble was drawn first without replacement. $\newline$ Since the first marble is not replaced, the total number of marbles decreases by one, and the number of green marbles decreases by one.Now we have $4$ blue marbles and $8$ $5$ green marbles left, making a total of $8$ $6$ marbles. $\newline$ The probability of drawing a blue marble on the second draw without replacement is now $8$ $7$ .
Calculate Probability without Replacement: The total number of marbles is $4$ blue + $8$ green = $12$ marbles. $\newline$ The probability of drawing a blue marble is the number of blue marbles divided by the total number of marbles, which is $\frac{4}{12}$ or $\frac{1}{3}$ .Using the conditional probability formula $P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$ , we need to find $P(\text{blue second} | \text{green first})$ . $\newline$ Since the first marble is replaced, the events are independent, and $P(A \text{ and } B) = P(A) \times P(B)$ . $\newline$ However, since we are given that a green marble is drawn first, $P(B)$ is certain ( $P(B) = 1$ ), and thus $8$ $0$ .Therefore, $P(\text{blue second} | \text{green first})$ with replacement is simply the probability of drawing a blue marble, which is $\frac{1}{3}$ or approximately $8$ $3$ . $\newline$ b. Calculate the probability of drawing a blue marble second given that a green marble was drawn first without replacement. $\newline$ Since the first marble is not replaced, the total number of marbles decreases by one, and the number of green marbles decreases by one.Now we have $4$ blue marbles and $8$ $5$ green marbles left, making a total of $8$ $6$ marbles. $\newline$ The probability of drawing a blue marble on the second draw without replacement is now $8$ $7$ .Therefore, $P(\text{blue second} | \text{green first})$ without replacement is $8$ $7$ , which is approximately $12$ $0$ .