To solve this problem, let's calculate $P(1 \leq X \leq 3)$ for a binomial random variable $X$ with parameters $n = 7$ and $p = 0.3$.

1. Identify the Probability Expression
   The probability $P(1 \leq X \leq 3)$ represents the sum of probabilities for $X = 1$, $X = 2$, and $X = 3$: $P(1 \leq X \leq 3) = P(X = 1) + P(X = 2) + P(X = 3)$

2. Use the Binomial Probability Formula
   The probability of observing $k$ successes in a binomial distribution is: $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$ where $\binom{n}{k}$ is the binomial coefficient, which can be calculated as: $\binom{n}{k} = \frac{n!}{k!(n - k)!}$

3. Calculate Each Probability
   Given $n = 7$ and $p = 0.3$, we calculate $P(X = 1)$, $P(X = 2)$, and $P(X = 3)$ individually.

   • For $P(X = 1)$: $P(X = 1) = \binom{7}{1} (0.3)^1 (0.7)^6$

   • For $P(X = 2)$: $P(X = 2) = \binom{7}{2} (0.3)^2 (0.7)^5$

   • For $P(X = 3)$: $P(X = 3) = \binom{7}{3} (0.3)^3 (0.7)^4$

4. Sum the Probabilities
   After calculating each term, sum them to find $P(1 \leq X \leq 3)$.

Let’s calculate this.The probability $P(1 \leq X \leq 3)$ for a binomial variable with $n = 7$ and $p = 0.3$ is approximately 0.79161.

Would you like more details on the steps or have any questions?

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Here are 5 related questions for practice:

1. What is the probability $P(X = 4)$ for the same binomial distribution?
2. How would you calculate $P(X \leq 3)$ for this distribution?
3. What is the probability $P(X = 0)$ in this case?
4. How would the probability change if $p$ were 0.5 instead of 0.3?
5. How do you interpret the expected value of a binomial distribution with these parameters?

Tip: When calculating binomial probabilities by hand, double-check binomial coefficients and powers to ensure precision.