Hi! I'm Maddie. I need some help on my homework. Here are three problems I need help with:
A bowl contains only red marbles, blue marbles and green marbles. The probability of selecting a red marble from the bowl is 3/13. The probability of selecting a blue marble from the bowl is 2/5. There are fewer than 100 marbles in the bowl. What is the probability of selecting, at random and without replacement, a green marble and then a red marble from the bowl on the first two selections? Express your answer as a common fraction.
The second one is:
The digits 2, 3, 4, 7, and 8 are each used once in a random order to form a five-digit number. What is the probability that the resulting number is divisible by 4? Express your answer as a common fraction.
And the last one is:
A six-sided die (with numbers 1 through 6) and an eight-sided die (with numbers 1 through 8) are rolled. What is the probability that there is exactly one 6 showing? Express your answer as a common fraction.
Thank you so much if you can help!! It would be amazing if you could help me out.
Ha!These are all from AoPS. Anyways,
1) probability of green then red = p(green) * p(red). Take a common multiple of 13 & 5 that's under 100 - I chose 65. The numerators all have to add up to the denominator in the probability. p(red) = 15/65. p(blue) = 26/65. so p(green) has to equal (65 - (15+26))/65 = 24/65. Using cross multiplication, you can find out that 24/65 * 15/65 = 9/104.
2) A number is divisible by 4 if and only if the number formed by its last two digits is divisible by 4. Using the given digits, we find that the only two-digit numbers that are divisible by 4 are 24, 28, 32, 48, 72, and 84.
The probability that the number is divisible by 4 is the number of last-2-digit combinations that make it divisible by four divided by the total ways to choose the last 2 digits.
There are 4 * 5 = 20 ways to choose the last two digits, and they are all equally likely, so the probability that the number is divisible by 4 is 6/20 = 3/10.
3) If there is exactly one 6 showing, then the 6 must be on the six-sided die or the eight-sided die (but not both).
The probability of getting a 6 with the six-sided die is 1/6, and the probability of not rolling a 6 with the eight-sided die is 7/8.
The probability of not getting a 6 with the six-sided die is 5/6, and the probability of rolling a 6 with the eight-sided die is 1/8.
Therefore, the probability that there is exactly one 6 is 1/6 * 7/8 + 5/6 * 1/8 = 7/48 + 5/48 = 12/48 = 1/4
So your answers are:
1) 9/104
2) 3/10
3) 1/4
Hi Maddie! I'd be happy to help you with your homework problems. Let's take them one by one:
Problem 1:
To find the probability of selecting a green marble and then a red marble without replacement, we need to multiply the probabilities of each event.
Given that there are fewer than 100 marbles in the bowl, we can assume there are a total of 99 marbles (since there are no other mentioned colors).
Let's denote the number of green marbles as G and the number of red marbles as R.
So, we have:
Probability of selecting a green marble on the first selection: G/99
Probability of selecting a red marble on the second selection: (R-1)/(99-1) = (R-1)/98 (since one green marble has been removed)
We're given that the probability of selecting a red marble is 3/13. This means that R/99 = 3/13, which we can cross-multiply and solve for R:
13R = 3 * 99
R = (3 * 99) / 13
R ≈ 22.85
Since the number of marbles must be a whole number, the closest whole number to 22.85 is 23. Therefore, R = 23.
Now we can calculate the probability:
Probability of selecting a green marble and then a red marble =
(G/99) * ((R-1)/98) =
(G/99) * (22/98) =
(G/99) * (11/49) =
11G/4851
So, the probability of selecting a green marble and then a red marble is 11G/4851.
Problem 2:
To find the probability that a five-digit number formed using the digits 2, 3, 4, 7, and 8 is divisible by 4, we need to consider the divisibility rule for 4, which states that a number is divisible by 4 if the last two digits of the number are divisible by 4.
Since the digits can be rearranged in a random order, we have to determine all possible combinations.
First, let's count the total number of possible five-digit numbers that can be formed using the given digits. Since there are 5 different digits, the total number of permutations is 5!.
5! = 5 * 4 * 3 * 2 * 1 = 120
Out of these 120 numbers, we need to determine how many are divisible by 4.
Numbers divisible by 4 must have their last two digits as one of the following:
- 24
- 28
- 32
- 48
- 72
- 84
We need to consider all possible arrangements of the remaining three digits (2, 3, 7). Since repetition is not allowed, we have 3! = 3 * 2 * 1 = 6 possible arrangements.
So, the total number of five-digit numbers divisible by 4 is 6 * 6 = 36.
The probability of selecting a number divisible by 4 is then:
36/120 = 3/10
Therefore, the probability that the resulting number is divisible by 4 is 3/10.
Problem 3:
To find the probability of getting exactly one 6 when rolling both dice, we need to consider all possible outcomes and count the favorable outcomes.
Let's consider the six-sided die (D6) and the eight-sided die (D8) separately.
For D6, there are 6 possible outcomes (1, 2, 3, 4, 5, 6) and we want exactly one 6. So, the probability of rolling exactly one 6 with D6 is 1/6.
For D8, there are 8 possible outcomes (1, 2, 3, 4, 5, 6, 7, 8) and we want to avoid rolling a 6. So, the probability of not rolling a 6 with D8 is 7/8.
To find the overall probability of getting exactly one 6 when rolling both dice, we multiply the probabilities of each die:
Probability of rolling exactly one 6 = (1/6) * (7/8) = 7/48
Therefore, the probability that there is exactly one 6 showing is 7/48.
I hope that helps! Let me know if you have any further questions.