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A bag contains $6$ red balls, $4$ green balls, and $3$ blue balls. If we choose a ball, then another ball without putting the first one back in the bag, what is the probability that the first ball will be green and the second will be red?

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Probability of independent and dependent events

Full solution

Q. A bag contains

6

red balls,

4

green balls, and

3

blue balls. If we choose a ball, then another ball without putting the first one back in the bag, what is the probability that the first ball will be green and the second will be red?

Calculate Total Number of Balls: Calculate the total number of balls in the bag. $\newline$ The bag contains $6$ red balls, $4$ green balls, and $3$ blue balls. So, the total number of balls is $6 + 4 + 3 = 13$ balls.
Calculate Probability of Drawing Green Ball: Calculate the probability of drawing a green ball first. $\newline$ Since there are $4$ green balls and the total number of balls is $13$ , the probability of drawing a green ball first is $\frac{4}{13}$ .
Calculate Number of Balls Left: Calculate the number of balls left after drawing one green ball. $\newline$ After one green ball is drawn, there are $13 - 1 = 12$ balls left in the bag.
Calculate Probability of Drawing Red Ball: Calculate the probability of drawing a red ball after drawing a green ball. Since there are still $6$ red balls left and now only $12$ balls in total, the probability of drawing a red ball after a green ball is $\frac{6}{12}$ .
Calculate Combined Probability: Calculate the combined probability of both events happening in sequence. $\newline$ To find the combined probability of drawing a green ball first and then a red ball, we multiply the probabilities of each individual event. So, the combined probability is $(\frac{4}{13}) \times (\frac{6}{12})$ .
Simplify Combined Probability: Simplify the combined probability. $\newline$ The combined probability simplifies to $(\frac{4}{13}) \times (\frac{1}{2}) = \frac{4}{26} = \frac{2}{13}$ after simplifying the fraction.

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