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A bag contains 8 blue and 7 green balls. 2 balls are drawn one by one without replacement. What is the probability that the balls are alternately of different colors?
- a)9/21
- b)8/17
- c)5/14
- d)8/15
- e)9/19
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A bag contains 8 blue and 7 green balls. 2 balls are drawn one by one ...
When 1st is blue, prob. = 8/15 * 7/14
When 1st is green, prob. = 7/15 * 8/14
Add both cases.
When 1st is green, prob. = 7/15 * 8/14
Add both cases.
Most Upvoted Answer
A bag contains 8 blue and 7 green balls. 2 balls are drawn one by one ...
To solve this problem, we can use the concept of conditional probability.
Let's break down the problem step by step:
Step 1: Calculate the probability of drawing a blue ball first.
There are 8 blue balls out of a total of 15 balls in the bag. Therefore, the probability of drawing a blue ball first is 8/15.
Step 2: Calculate the probability of drawing a green ball second, given that a blue ball was drawn first.
After drawing a blue ball, there are now 7 blue balls and 7 green balls left in the bag. Therefore, the probability of drawing a green ball second is 7/14.
Step 3: Calculate the probability of drawing a green ball first.
There are 7 green balls out of a total of 15 balls in the bag. Therefore, the probability of drawing a green ball first is 7/15.
Step 4: Calculate the probability of drawing a blue ball second, given that a green ball was drawn first.
After drawing a green ball, there are now 8 blue balls and 6 green balls left in the bag. Therefore, the probability of drawing a blue ball second is 8/14.
Step 5: Calculate the overall probability of drawing alternately different colored balls.
Since there are two possible scenarios (blue-green or green-blue), we need to calculate the probability for each scenario and then add them together.
Scenario 1: Probability of drawing a blue ball first and then a green ball second = (8/15) * (7/14) = 56/210
Scenario 2: Probability of drawing a green ball first and then a blue ball second = (7/15) * (8/14) = 56/210
Finally, we add the probabilities of both scenarios:
56/210 + 56/210 = 112/210 = 8/15
Therefore, the probability that the balls are alternately of different colors is 8/15, which corresponds to option D.
Let's break down the problem step by step:
Step 1: Calculate the probability of drawing a blue ball first.
There are 8 blue balls out of a total of 15 balls in the bag. Therefore, the probability of drawing a blue ball first is 8/15.
Step 2: Calculate the probability of drawing a green ball second, given that a blue ball was drawn first.
After drawing a blue ball, there are now 7 blue balls and 7 green balls left in the bag. Therefore, the probability of drawing a green ball second is 7/14.
Step 3: Calculate the probability of drawing a green ball first.
There are 7 green balls out of a total of 15 balls in the bag. Therefore, the probability of drawing a green ball first is 7/15.
Step 4: Calculate the probability of drawing a blue ball second, given that a green ball was drawn first.
After drawing a green ball, there are now 8 blue balls and 6 green balls left in the bag. Therefore, the probability of drawing a blue ball second is 8/14.
Step 5: Calculate the overall probability of drawing alternately different colored balls.
Since there are two possible scenarios (blue-green or green-blue), we need to calculate the probability for each scenario and then add them together.
Scenario 1: Probability of drawing a blue ball first and then a green ball second = (8/15) * (7/14) = 56/210
Scenario 2: Probability of drawing a green ball first and then a blue ball second = (7/15) * (8/14) = 56/210
Finally, we add the probabilities of both scenarios:
56/210 + 56/210 = 112/210 = 8/15
Therefore, the probability that the balls are alternately of different colors is 8/15, which corresponds to option D.
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