We are asked to find $P(X < 2)$ for a binomial distribution with parameters $n = 5$ (number of trials) and $p = 0.7$ (probability of success).

Steps:

The binomial probability $P(X = k)$ is given by the formula:

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

where:

• $\binom{n}{k}$ is the binomial coefficient, calculated as $\frac{n!}{k!(n-k)!}$,
• $p$ is the probability of success on each trial,
• $n$ is the number of trials,
• $k$ is the number of successes.

We need to find $P(X < 2)$, which is the probability that $X$ is less than 2. This means we are looking for $P(X = 0)$ and $P(X = 1)$, and then summing them.

1. Find $P(X = 0)$:

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

2. Find $P(X = 1)$:

$P(X = 1) = \binom{5}{1} (0.7)^1 (0.3)^4 = 5 \times 0.7 \times 0.0081 = 0.02835$

3. Find $P(X < 2)$:

Now, we can sum these probabilities to get $P(X < 2)$:

$P(X < 2) = P(X = 0) + P(X = 1) = 0.00243 + 0.02835 = 0.03078$

Final Answer:

$P(X < 2) = 0.0308$

So, the probability that $X$ is less than 2 is approximately 0.0308.

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Would you like more details or have any questions?

Here are some related questions for further exploration:

1. How would the probability change if $p$ was decreased to 0.5?
2. What is the probability of exactly 3 successes (i.e., $P(X = 3)$)?
3. How can you generalize the binomial probability formula for larger values of $n$?
4. How would the answer change if the number of trials increased to 10?
5. What happens to the binomial distribution as the number of trials $n$ becomes very large?

Tip: When calculating probabilities for binomial distributions, it's often useful to use a calculator or statistical software to avoid manual calculation errors.