SOLUTION: A bag contains 25 chips numbered 1 to 25. If a chip is drawn randomly from the bag, what
is the probability that it is
a 7 or 15?
b. 5 or a number divisible by 3?
c. ev
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Question 1192038: A bag contains 25 chips numbered 1 to 25. If a chip is drawn randomly from the bag, what
is the probability that it is
a 7 or 15?
b. 5 or a number divisible by 3?
c. even or divisible by 3?
d a number divisible by 3 or divisible by 4?
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
Answers:
a) 2/25
b) 9/25
c) 16/25
d) 12/25
===================================================
Explanation on how I got those answers:
Part (a)
There are 2 chips we want out of 25 total.
Therefore, 2/25 is the probability of getting either a "7" or a "15".
-------------------------
Part (b)
Event space = {3, 5, 6, 9, 12, 15, 18, 21, 24}
The event space is the set of stuff we want to happen, which is getting a multiple of 3 or getting "5".
There are 9 numbers in that list.
This is out of 25 total items in the sample space.
That's why 9/25 is the answer.
-------------------------
Part (c)
A = number is even = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24}
B = number is divisible by 3 = {3, 6, 9, 12, 15, 18, 21, 24}
Each item in either set above must be between 1 and 25.
C = the number is even, or divisible by 3, or both
C = A union B
C = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24}
All of the items in set C represent stuff we want to happen.
There are 16 values in set C out of 25 values total.
Answer: 16/25
-------------------------
Part (d)
A = number is divisible by 3 = {3, 6, 9, 12, 15, 18, 21, 24}
B = number is divisible by 4 = {4, 8, 12, 16, 20, 24}
C = the number is divisible by 3, or divisible by 4, or both
C = A union B
C = {3, 4, 6, 8, 9, 12, 15, 16, 18, 20, 21, 24}
There are 12 items in set C out of 25 total.
The probability of getting a number divisible by 3, divisible by 4, or both is 12/25
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