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Answer:
[tex]1. \: P(7 \text{ or } 15) = \frac{2}{15} [/tex]
[tex]2. \: P(5 \text{ or divisible by } 3) = \frac{2}{5} [/tex]
[tex]3. \: P(\text{even or divisible by } 3) = \frac{2}{3} [/tex]
[tex]4. \: P(\text{divisible by } 3 \text{ or } 4) = \frac{7}{15}[/tex]
Step-by-step explanation:
Total Number of Outcomes
The total number of chips in the bowl is 15. Therefore, the total number of outcomes is:
N = 15.
Step 1: Probability of drawing a chip that is 7 or 15
The favorable outcomes for this condition are the chips numbered 7 and 15. Thus, the number of favorable outcomes is:
[tex]N_{7 \text{ or } 15} = 2.[/tex]
[tex]P(7 \text{ or } 15) = \frac{N_{7 \text{ or } 15}}{N} = \frac{2}{15}.[/tex]
Step 2: Probability of drawing a chip that is 5 or a number divisible by 3
The favorable outcomes for this condition include the chip numbered 5 and the chips divisible by 3. The numbers divisible by 3 from 1 to 15 are:
- 5
- 3, 6, 9, 12, 15 (5 numbers)
[tex]N_{5 \text{ or divisible by } 3} = 1 + 5 = 6.[/tex]
[tex]{P(5 \text{ or divisible by } 3) = \frac{N_{5 \text{ or divisible by } 3}}{N} = \frac{6}{15} = \frac{2}{5}.}[/tex]
Step 3: Probability of drawing a chip that is even or divisible by 3
The even numbers from 1 to 15 are:
[tex]2, 4, 6, 8, 10, 12, 14 \quad (7 \text{ even \: numbers}).[/tex]
[tex]3, 6, 9, 12, 15 \quad (5 \text{ numbers}).[/tex]
Now, we need to count the unique outcomes. The even numbers and the numbers divisible by 3 are:
- Even: 2, 4, 6, 8, 10, 12, 14
- Divisible by 3: 3, 6, 9, 12, 15
The common numbers are 6 and 12. Thus, the total unique favorable outcomes are:
[tex]{7 \text{ (even)} + 5 \text{ (divisible by 3)} - 2 \text{ (common)} = 10.}[/tex]
[tex]{P(\text{even or divisible by } 3) = \frac{10}{15} = \frac{2}{3}.}[/tex]
Step 4: Probability of drawing a chip that is divisible by 3 or divisible by 4
[tex]3, 6, 9, 12, 15 \quad (5 \text{ numbers}).[/tex]
[tex]4, 8, 12 \quad (3 \text{ numbers}).[/tex]
The common number is 12. Thus, the total unique favorable outcomes are:
[tex]{5 \text{ (divisible by 3)} + 3 \text{ (divisible by 4)} - 1 \text{ (common)} = 7.}[/tex]
[tex]P(\text{divisible by } 3 \text{ or } 4) = \frac{7}{15}.[/tex]