Answer:
Step-by-step explanation:
Since there are 2 red and 5 white balls in the box, the total number of balls in the bag = 2+5 = 7balls.
Probability that at least 1 ball was red, given that the first ball was replaced before the second can be calculated as shown;
Since at least 1 ball picked at random, was red, this means the selection can either be a red ball first then a white ball or two red balls.
Probability of selecting a red ball first then a white ball with replacement = (2/7*5/7) = 10/49
Probability of selecting two red balls with replacement = 2/7*2/7 = 4/49
The probability that at least 1 ball was red given that the first ball was replaced before the second draw= 10/49+4/49 = 14/49
If the balls were not replaced before the second draw
Probability of selecting a red ball first then a white ball without replacement = (2/7*5/6) = 10/42 = 5/21
Probability of selecting two red balls without replacement = 2/7*2/6 = 4/42 = 2/21
The probability that at least 1 ball was red given that the first ball was not replaced before the second draw = 5/21+4/21 = 9/21 = 3/7
The probability that at least 1 ball was red, given that the first ball was replaced before the second draw is 28.5%; and the probability that at least 1 ball was red, given that the first ball was not replaced before the second draw is 22.5%.
Since two balls are drawn in succession out of a box containing 2 red and 5 white balls, to find the probability that at least 1 ball was red, given that the first ball was A) replaced before the second draw; and B) not replaced before the second draw; the following calculations must be performed:
Therefore, the probability that at least 1 ball was red, given that the first ball was replaced before the second draw is 28.5%; and the probability that at least 1 ball was red, given that the first ball was not replaced before the second draw is 22.5%.
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