a number is picked from the set (1,2,3,_,18).find the probability that it is either a multiple 3 or a number less than 7. find an answer it
answered in a related question below.
Actually looking back at my answer in the Related Question, I see that I made an error
The logical connector was OR , so
multiples of 3 : 3,6,9,12,15,18
or
less than 7: 1,2,3,4,5,6
can't count the 3 and 6 twice, so there are 10 such numbers
prob = 10/18 = 5/9
To find the probability, we need to determine the number of favorable outcomes (numbers that are either multiples of 3 or less than 7) and divide it by the total number of possible outcomes.
First, let's determine the number of favorable outcomes:
1. Multiples of 3: In the given set, we have the multiples of 3 as 3, 6, 9, 12, 15, and 18. So, there are 6 numbers that are multiples of 3.
2. Numbers less than 7: From 1 to 6 (inclusive), we have 6 numbers that are less than 7.
Next, let's determine the total number of possible outcomes:
Given the set (1, 2, 3, _, 18), there are 18 - 3 + 1 = 16 numbers in total.
Now, we need to add the number of multiples of 3 and numbers less than 7 together, but we need to make sure we don't double-count the numbers that satisfy both conditions.
The numbers that satisfy both conditions (multiples of 3 and less than 7) are 3 and 6.
So, we have:
- 6 multiples of 3 (excluding 3 and 6)
- 6 numbers less than 7 (excluding 3 and 6)
- 2 numbers satisfying both conditions (3 and 6)
Adding them together: 6 + 6 - 2 = 10
Therefore, the number of favorable outcomes is 10.
Finally, the probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes:
Probability = Number of favorable outcomes / Total number of possible outcomes
= 10 / 16
= 5 / 8
= 0.625 (or 62.5%)
So, the probability that a number picked from the set is either a multiple of 3 or a number less than 7 is 0.625 or 62.5%.