A box contains four red balls and two white balls. A trialconsists of mixing up

Question

A box contains four red balls and two white balls. A trialconsists of mixing up the balls, selecting a ball at random, notingits color and then returning it to the box. So: 1) What is the probability of obtaining a red ball in onetrial? 2) For two trials, find the probability that a red ball iselected on the first trial and a white ball is selected on thesecond trial. 3) For two trials, find the probability that a white ball isselected in the first trial and a red ball in the secondtrial. 4) For two trials, find that probability that the number ofred balls selected is one. 5) For three trials, find the probability that two red ballsare selected. 6) For three trials, find the probability that the number ofred balls selected is not greater than two A box contains four red balls and two white balls. A trialconsists of mixing up the balls, selecting a ball at random, notingits color and then returning it to the box. So: 1) What is the probability of obtaining a red ball in onetrial? 2) For two trials, find the probability that a red ball iselected on the first trial and a white ball is selected on thesecond trial. 3) For two trials, find the probability that a white ball isselected in the first trial and a red ball in the secondtrial. 4) For two trials, find that probability that the number ofred balls selected is one. 5) For three trials, find the probability that two red ballsare selected. 6) For three trials, find the probability that the number ofred balls selected is not greater than two

Explanation / Answer

1) prob = 4/6 = 2/3 2) prob = (4/6)(2/6) = 2/9 3) prob = (2/6)(4/6) = 2/9 4) We can have Red in first trial + White in 2nd trial, orWhite in first trial + Red in 2nd trial prob = P(Red-White) + P(White-Red) = (4/6)(2/6) + (2/6)(4/6) =4/9 5) prob = (4/6)(4/6) = 4/9 6) P(not greater than two) = 1 - P(greater than two) = 1 - P(3red balls) = 1 - (4/6)(4/6)(4/6) = 19/27 Hope this helps

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