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.
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.
1 answer
3 answers