Tickets numbered from one to 40 are mixed up and one is drawn.
What is the probability that number is divisible by 3 or 5?
What is the probability that number is divisible by 3 and 5?
8/40+13/40-2/40 =19/40
8/40 numbers divisible by 5
13/40 numbers divisible by 3
2/40 overlapping numbers which are 15 and 30
n = 40
n(A) = 13+8-2 = 21-2
⇒ P(A) = 19/40
Yes, that is correct for the probability that the number drawn is divisible by 3 or 5.
For the probability that the number is divisible by 3 and 5, we need to find the number of multiples of 15 in the range of 1 to 40.
Multiples of 15 are: 15, 30. So, there are 2 multiples of 15 in the range of 1 to 40.
Therefore, the probability that the number drawn is divisible by both 3 and 5 (i.e., by 15) is:
P(B) = 2/40
Alternatively, we can use the formula for the probability of the intersection of two events:
P(A ∩ B) = P(A) × P(B|A)
Here, P(A) is the probability of the event that the number is divisible by 3, which is 13/40.
The probability of the event that the number is divisible by 5, given that it is already known to be divisible by 3 (i.e., given the event A), is 2/13, because the only multiples of 5 that are also multiples of 3 are 15 and 30.
So, we can calculate P(B) as:
P(B) = P(A ∩ B) / P(A|B) = (13/40) × (2/13) = 2/40
Either way, we get the same answer: P(B) = 2/40 = 1/20.
Well, let's break it down!
Question 1: What is the probability that the number drawn is divisible by 3 or 5?
To find the probability of an event happening, we need to determine the number of favorable outcomes divided by the total number of possible outcomes.
In this case, let's figure out the number of numbers from one to 40 that are divisible by 3 or 5.
We have: 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30, 33, 35, 36, 39, and 40.
Counting them all up, we got 19 numbers. So the number of favorable outcomes is 19.
Since there are 40 total tickets, the total number of possible outcomes is 40.
Now, we can calculate the probability: 19/40 = 0.475.
So the probability is approximately 0.475, or around 47.5%.
Question 2: What is the probability that the number drawn is divisible by both 3 and 5?
Well, in this case, we need to find the numbers that are divisible by both 3 and 5.
The only number in our list is 15. So we have only one favorable outcome.
Again, the total number of possible outcomes is 40.
Calculating the probability: 1/40 = 0.025.
So the probability is approximately 0.025, or 2.5%.
Remember folks, the probability of picking the winning ticket is always a bit of a long shot!
To find the probability of an event occurring, we divide the number of favorable outcomes by the total number of possible outcomes.
Let's start by counting the total number of tickets in the sample space, which is 40.
1. Probability that the number is divisible by 3 or 5:
To find the number of favorable outcomes, we need to count the tickets that are divisible by 3 or 5. We can break down the problem into two parts:
a) Numbers divisible by 3: There are 13 numbers (3, 6, 9, 12, ..., 39) that are divisible by 3.
b) Numbers divisible by 5: There are 8 numbers (5, 10, 15, ..., 40) that are divisible by 5.
However, we need to be careful not to count any numbers twice. To avoid that, let's find the numbers that are divisible by both 3 and 5, and subtract them from our total count.
c) Numbers divisible by both 3 and 5: There is only one number that is divisible by both 3 and 5, which is 15.
Hence, the number of favorable outcomes is 13 + 8 - 1 = 20.
Therefore, the probability that the number is divisible by 3 or 5 is 20/40 = 1/2.
2. Probability that the number is divisible by 3 and 5:
From the previous calculation, we found that there is only one number, 15, that satisfies both conditions (divisible by 3 and 5).
Therefore, the number of favorable outcomes is 1.
Hence, the probability that the number is divisible by 3 and 5 is 1/40.
x: How many multiples of 3 are there in 1-40?
y: How many multiples of 5?
z: how many multiples of 15?
P(x or y) = (x+y-z)/40
P(x and y) = P(z) = z/40