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.
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.
To solve both of these questions, we can use the concept of probability.
In the first question, we have four draws with replacement from the box. This means that after each draw, the selected number is placed back in the box before the next draw is made. To find the probability of drawing at least one "2," we can look at the complementary event - the event of not drawing any "2"s.
Since there are five numbers in the box (1, 2, 2, 3, 3) and we have four draws, the total number of possible outcomes is 5^4, which is 625. To calculate the number of outcomes without a "2," we need to consider all the other remaining numbers (1 and 3) and the fact that we have four draws. So, there are 2^4 = 16 outcomes without a "2".
Therefore, the probability of not drawing any "2"s is 16/625. To find the probability of drawing at least one "2," we subtract this probability from 1.
1 - 16/625 = 609/625 ≈ 0.9744
So, the probability of drawing at least one "2" is approximately 0.9744.
For the second question, we have four draws without replacement. This means that after each draw, the selected number is not placed back in the box before the next draw is made.
To find the probability of drawing at least one "2," we again look at the complementary event of not drawing any "2"s.
Since we have four draws, the total number of possible outcomes is reduced. In the first draw, there are five numbers to choose from. In the second draw, there are only four remaining numbers, and so on. Therefore, the total number of possible outcomes without a "2" is 5 * 4 * 3 * 2.
To find the probability, we need to divide the number of outcomes without a "2" by the total number of possible outcomes.
(5 * 4 * 3 * 2) / (5 * 4 * 3 * 2 * 1) = 1/5 = 0.20
To express this probability as a percentage, we multiply it by 100.
0.20 * 100 = 20
So, the probability of drawing at least one "2" is 20%.
I hope this explanation helps!
To calculate the probability of drawing a "2" at least once, we can use the complement rule.
1. Probability with replacement:
The probability of drawing a "2" on a single draw is 2 out of 5 since there are two "2" in the box. Therefore, the probability of not drawing a "2" on a single draw is 3 out of 5. Since the draws are made with replacement, the probabilities are independent of each other. To find the probability of not drawing a "2" in all four draws, we can multiply the probabilities of not drawing a "2" in each draw: (3/5)^4.
Using the complement rule, the probability of drawing a "2" at least once is 1 - (3/5)^4.
2. Probability without replacement:
The probability of drawing a "2" on the first draw is 2 out of 5. After removing one "2" from the box, the probability of drawing a "2" on the second draw is 1 out of 4. Similarly, the probabilities for the third and fourth draws are 1 out of 3 and 1 out of 2, respectively. To find the probability of drawing a "2" at least once, we can calculate the probability of not drawing a "2" in all four draws and subtract it from 1: 1 - (3/5) * (2/4) * (2/3) * (1/2).
Now, let's calculate the answers:
1. Probability with replacement:
1 - (3/5)^4 ≈ 0.8704
2. Probability without replacement:
1 - (3/5) * (2/4) * (2/3) * (1/2) ≈ 0.7333
Therefore:
- The probability of drawing a "2" at least once with replacement is approximately 0.8704 (rounded to 4 decimal places).
- The probability of drawing a "2" at least once without replacement is approximately 73% (rounded to the nearest whole percent).