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A bag contains 66 red balls, 44 green balls, and 33 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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Q. A bag contains 66 red balls, 44 green balls, and 33 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?
  1. Calculate Total Number of Balls: Calculate the total number of balls in the bag.\newlineThe bag contains 66 red balls, 44 green balls, and 33 blue balls. So, the total number of balls is 6+4+3=136 + 4 + 3 = 13 balls.
  2. Calculate Probability of Drawing Green Ball: Calculate the probability of drawing a green ball first.\newlineSince there are 44 green balls and the total number of balls is 1313, the probability of drawing a green ball first is 413\frac{4}{13}.
  3. Calculate Number of Balls Left: Calculate the number of balls left after drawing one green ball.\newlineAfter one green ball is drawn, there are 131=1213 - 1 = 12 balls left in the bag.
  4. Calculate Probability of Drawing Red Ball: Calculate the probability of drawing a red ball after drawing a green ball. Since there are still 66 red balls left and now only 1212 balls in total, the probability of drawing a red ball after a green ball is 612\frac{6}{12}.
  5. Calculate Combined Probability: Calculate the combined probability of both events happening in sequence.\newlineTo 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 (413)×(612)(\frac{4}{13}) \times (\frac{6}{12}).
  6. Simplify Combined Probability: Simplify the combined probability.\newlineThe combined probability simplifies to (413)×(12)=426=213(\frac{4}{13}) \times (\frac{1}{2}) = \frac{4}{26} = \frac{2}{13} after simplifying the fraction.

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