Sagot :
Answer:
a. [tex]\frac{2}{15}[/tex]
b. [tex]\frac{2}{5}[/tex]
c. [tex]\frac{2}{3}[/tex]
d. [tex]\frac{7}{15}[/tex]
Step-by-step explanation:
Probabilities with multiple events fall on the category of compound events. Compound events are either mutually inclusive, meaning, they can occur at the same time, or mutually exclusive, meaning, they cannot occur at the same time.
An example of a mutually inclusive event is rolling a dice, and expecting 2 results, an even result, or 4. When you get a 4, you satisfy both events. The 2 events are mutually inclusive.
An example of a mutually exclusive event is rolling a dice, and expecting 2 results, an even result, or an odd number. Those 2 events will never happen at the same time. The 2 events are mutually exclusive.
The probability of 2 compound events, A and B, is given by the formula
[tex]P (A \ or\ B) = P(A) + P(B) - P(A \ and\ B)[/tex]
where
P (A or B) is the probability that A or B occurs.
P(A) is the probability of A occurring.
P(B) is the probability of B occurring.
P(A and B ) is the probability that A and B occur at the same time.
We need to subtract the instance when 2 events occur at the same time to avoid double counting. If 2 events don't happen at the same time, or are mutually exclusive, P(A and B) is 0.
For this problem, our total number of outcomes is always 15, since there are 15 chips in the bowl.
a. 7 or 15?
Our favorable outcomes are 7 or 15. The events cannot happen at the same time. They are mutually exclusive. P(A and B) is 0.
[tex]P (7 \ or\ 15) = P(7) + P(15) - P(7 \ and\ 15)\\P (7 \ or\ 15) = \frac{1}{15} + \frac{1}{15} - 0\\\\P (7 \ or\ 15) = \frac{2}{15}[/tex]
The probability of drawing 7 or 15 is[tex]\frac{2}{15}[/tex].
b. 5 or a number divisible by 3?
The multiples of 3 from 1 to 15 are as follows: 3, 6, 9, 12, 15. We have 5 favorable outcomes for a number to be divisible by 3.
We have 1 favorable outcome for a number to be 5. The events are mutually exclusive. P(A and B) is 0.
[tex]P (5 \ or\ div. \ by\ 3) = P(5) + P(div. \ by \ 3) - P(5 \ and \ div. \ by \ 3)\\\\P (5 \ or\ div. \ by\ 3) = \frac{1}{15} + \frac{5}{15}- 0\\\\P (5 \ or\ div. \ by\ 3) = \frac{6}{15}\\\\P (5 \ or\ div. \ by\ 3) = \frac{2}{5}\\[/tex]
The probability of drawing 5 or a number divisible by 3 is [tex]\frac{2}{15}[/tex].
c. even or divisible by 3?
The even numbers from 1 to 15 are as follows: 2, 4, 6, 8, 10, 12, 14. We have 7 favorable outcomes for a number to be even.
The multiples of 3 from 1 to 15 are as follows: 3, 6, 9, 12, 15. We have 5 favorable outcomes for a number to be divisible by 3.
As you can see, a number can both be divisible by 3 and also be even. These results are 6, 12. The events are mutually inclusive with 2 favorable outcomes.
If the result of the draw is 6, we satisfy both conditions. We need to eliminate 1 of those to avoid double counting.
[tex]P (even \ or \ div. \ by \ 3) = P(even) + P(div. by 3) - P(even \ and \ div. \ by \ 3)\\\\P (even \ or \ div. \ by \ 3) = \frac{7}{15} + \frac{5}{15} - \frac{2}{15}\\\\P (even \ or \ div. \ by \ 3) = \frac{10}{15}\\\\P (even \ or \ div. \ by \ 3) = \frac{2}{3}[/tex]
The probability of drawing an even number or a number divisible by 3 is [tex]\frac{2}{3}[/tex].
d. a number divisible by 3 or divisible by 4?
The multiples of 3 from 1 to 15 are as follows: 3, 6, 9, 12, and 15. We have 5 favorable outcomes for a number to be divisible by 3.
The multiples of 4 from 1 to 15 are as follows: 4, 8, and 12. We have 3 favorable outcomes for a number to be divisible by 3.
A number can both be divisible by 4 and 3. That occurs when the number is 12. The 2 events are mutually inclusive with 1 favorable outcome.
[tex]P (div. \ by \ 3 \ or \ div. \ by \ 4) = P(div. \ by \ 3) + P(div. \ by \ 4) - P(div. \ by \ 3 \ and \ div. \ by \ 4)\\\\P (div. \ by \ 3 \ or \ div. \ by \ 4) = \frac{5}{15} + \frac{3}{15} -\frac{1}{15}\\\\P (div. \ by \ 3 \ or \ div. \ by \ 4) = \frac{7}{15}[/tex]
The probability of drawing a number divisible by 3 or divisible by 4 is [tex]\frac{7}{15}[/tex].
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