For families with 5 children, let X be the number of children with Genetic Condition B. Can the following table be a probability distribution for the random variable X?

xP(x)

1

0.2639

2

0.2605

3

0.1585

4

0.0985

5

0.2186




yes

no

To determine whether the given table can represent a probability distribution for the random variable \( X \), the following two conditions must be satisfied:

1. **All probabilities must be non-negative**: Each value of \( P(x) \) must satisfy \( 0 \leq P(x) \leq 1 \).
2. **The sum of the probabilities must equal 1**: That is, \( \sum P(x) = 1 \).

Let's check these conditions.

### 1. Non-negativity check:
- \( P(1) = 0.2639 \)
- \( P(2) = 0.2605 \)
- \( P(3) = 0.1585 \)
- \( P(4) = 0.0985 \)
- \( P(5) = 0.2186 \)

Since all of these values are between 0 and 1, the first condition is satisfied.

### 2. Sum of probabilities:
Let's add the given probabilities to check if they sum to 1:
\[
0.2639 + 0.2605 + 0.1585 + 0.0985 + 0.2186 = 0.999999
\]

The sum is approximately 1, so the second condition is satisfied as well.

### Conclusion:
Yes, the table can represent a probability distribution for the random variable \( X \).

Would you like more details on probability distributions or have any other questions?

Here are 5 related questions to expand on this topic:
1. How can we calculate the mean of a probability distribution?
2. What is the variance of a probability distribution and how is it computed?
3. What is the difference between a probability mass function and a probability density function?
4. How do cumulative distribution functions relate to probability distributions?
5. What is the expected value and how does it relate to real-world scenarios?

**Tip:** Always check both the non-negativity and the sum of probabilities when verifying a probability distribution.

For families with 5 children, let X be the number of children with Genetic Condition B. Can the following table be a probability distribution for the random variable X?

x  |  P(x)
1  |  0.2639
2  |  0.2605
3  |  0.1585
4  |  0.0985
5  |  0.2186

Yes or No

Determine whether the given table can be a valid probability distribution for the random variable X, representing the number of children with Genetic Condition B in a family of 5. The solution involves checking non-negativity and sum of probabilities, making it suitable for Grades 9-12 students learning about probability distributions.

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