To find $P(X > 1)$ where $X$ follows a binomial distribution with $n = 5$ trials and probability of success $p = 0.7$, we can break it down step by step.

The binomial probability mass function (PMF) is given by:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

Where:

• $n = 5$ (the number of trials),
• $p = 0.7$ (the probability of success),
• $k$ is the number of successes.

We are asked to find $P(X > 1)$, which is the probability that the number of successes is greater than 1. This can be expressed as:

$P(X > 1) = 1 - P(X \leq 1)$

To calculate this, we need to first compute $P(X \leq 1)$, which includes $P(X = 0)$ and $P(X = 1)$:

$P(X = 0) = \binom{5}{0} (0.7)^0 (0.3)^5 = (1) \times (1) \times (0.3)^5 = 0.00243$ $P(X = 1) = \binom{5}{1} (0.7)^1 (0.3)^4 = 5 \times (0.7) \times (0.3)^4 = 0.02835$

Now, sum these probabilities:

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.00243 + 0.02835 = 0.03078$

Finally, to find $P(X > 1)$, subtract from 1:

$P(X > 1) = 1 - 0.03078 = 0.9692$

Thus, the probability $P(X > 1)$ is approximately 0.9692.

---

Would you like further details on binomial distribution? Here are some related questions:

1. How do you calculate the mean and variance of a binomial distribution?
2. What is the cumulative distribution function (CDF) for a binomial distribution?
3. How would the probability change if the number of trials or probability of success were different?
4. How do you apply the binomial distribution in real-world scenarios?
5. How can you use normal approximation to estimate binomial probabilities when $n$ is large?

Tip: If you're dealing with large $n$ values in a binomial distribution, consider using the normal approximation for quicker calculations.