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Sagot :
Step-by-step explanation:
To find the probability that a ball drawn from the bag is either red or even, we need to use the principle of counting and probability.
First, let's identify all the relevant information:
1. **Total number of balls**:
There are 7 red balls and 3 white balls, so the total number of balls is:
\[
7 + 3 = 10
\]
2. **Red balls**:
The red balls are numbered: 2, 4, 5, 6, 7, 8, 10.
3. **White balls**:
The white balls are numbered: 1, 3, 9.
4. **Even numbered balls**:
The even numbers among all the balls are: 2, 4, 6, 8, 10.
Next, we will find the probability of each event:
- **Probability of drawing a red ball (P(Red))**:
The number of red balls is 7.
\[
P(\text{Red}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{7}{10}
\]
- **Probability of drawing an even-numbered ball (P(Even))**:
The number of even-numbered balls is 5 (2, 4, 6, 8, 10).
\[
P(\text{Even}) = \frac{\text{Number of even-numbered balls}}{\text{Total number of balls}} = \frac{5}{10} = \frac{1}{2}
\]
- **Probability of drawing a ball that is both red and even (P(Red ∩ Even))**:
The red balls that are also even are: 2, 4, 6, 8, 10 (5 balls).
\[
P(\text{Red and Even}) = \frac{\text{Number of red and even balls}}{\text{Total number of balls}} = \frac{5}{10} = \frac{1}{2}
\]
To find the probability of drawing a ball that is either red or even, we use the formula for the union of two events:
\[
P(\text{Red or Even}) = P(\text{Red}) + P(\text{Even}) - P(\text{Red and Even})
\]
Substituting the values:
\[
P(\text{Red or Even}) = \frac{7}{10} + \frac{5}{10} - \frac{5}{10} = \frac{7}{10}
\]
Therefore, the probability that the ball drawn is either red or even is:
\[
P(\text{Red or Even}) = \frac{7}{10} = 0.7
\]
To express this probability as a percentage, we multiply by 100:
\[
0.7 \times 100 = 70\%
\]
Thus, the probability that a ball drawn is either red or even is \( 0.700 \) or 70.0%.
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