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I just asked a question a few minutes ago. But this is the last two questions I need help with. I have tried a couple different answers for both and still can't seem to get the correct ones.

Four draws are going to be made at random with replacement from the box

1 2 2 3 3

What is the probability that "2" is drawn at least once? Give your answer as a decimal rounded to 4 places.

Four draws are going to be made at random without replacement from the box

1 2 2 3 3

What is the probability that "2" is drawn at least once? Give your answer as a percentage rounded to a whole percent.

I thought the first answer was .0256, although I have tried it a couple different ways and every answer I got was wrong. For the second question I got 10% or 5% but I don't think that those are the answers either.

Thank you so much for any help! I keep looking over the videos and notes my teacher posted but these two questions make almost no sense to me since I keep getting them wrong.

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1 answer
  1. In the first, there is replacement, so the events are independent
    to get at least one of the 2's, we simply exclude the case of "no 2"
    prob(a 2) = 2/5, so prob(no 2) = 3/5
    prob(at least one 2 in 4 tries) = 1 - (3/5)^4 = 175/256 = appr .6836

    In the 2nd there is no replacement , a bit tricky
    since in 4 tries we are going to hit a 2 sometime
    do it by cases:
    exactly one 2:
    one possibility: 2133, which can be arranged in 4!/2! ways or 12 ways
    exactly two 2's:
    2213 --> arrange in 4!/2! way or 12
    2233 --> arrange in 4!/(2!2!) = 6 ways
    can't have more than two 2's
    total = 12+12+6 = 30
    total number of ways to arrange 12233 = 5!/(2!2!) = 30
    prob(at least 2, with replacement) = 30/30 = 1
    as said before, we must hit a 2 sometime.

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