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a bag contains 9 green marbles, 3 red marbles, and 8 blue marbles. one …

Question

a bag contains 9 green marbles, 3 red marbles, and 8 blue marbles. one marble is taken from the bag and put back after checking its color. a second marble is then taken out. what is the probability that the first is blue and the second green? a \\( \frac { 9 } { 50 } \\) b \\( \frac { 17 } { 20 } \\) c \\( \frac { 1 } { 19 } \\) d \\( \frac { 18 } { 95 } \\)

Explanation:

Step1: Calculate total number of marbles

Total marbles = \(9 + 3+8=20\)

Step2: Calculate probability of first - marble being blue

Probability of blue \(P(B)=\frac{8}{20}\) (since there are 8 blue marbles out of 20 total)

Step3: Calculate probability of second - marble being green

Since the first marble is put back, total marbles remain 20. Probability of green \(P(G)=\frac{9}{20}\) (since there are 9 green marbles out of 20 total)

Step4: Calculate the joint - probability

Since the two events (first - draw and second - draw) are independent (because of replacement), the joint - probability \(P = P(B)\times P(G)\)

$$ LATEXBLOCK0 $$

Answer:

A. \(\frac{9}{50}\)