Assume the random variable $X$ has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places.
\[
P(X> 3), n=6, p=0.5
\]
Answer
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Step 1 ：The question is asking for the probability that a binomial random variable $X$ with parameters $n=6$ and $p=0.5$ is greater than 3. This is equivalent to finding $1 - P(X \leq 3)$.

Step 2 ：The probability mass function of a binomial distribution is given by: $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$ where $\binom{n}{k}$ is the binomial coefficient, $p$ is the probability of success, and $n$ is the number of trials.

Step 3 ：We can calculate $P(X \leq 3)$ by summing up the probabilities $P(X=k)$ for $k=0,1,2,3$ and then subtract this from 1 to find $P(X>3)$.

Step 4 ：Given $n = 6$ and $p = 0.5$, we find that $P(X \leq 3) = 0.65625$.

Step 5 ：Subtracting this from 1, we find that $P(X>3) = 1 - P(X \leq 3) = 1 - 0.65625 = 0.3438$.

Step 6 ：Final Answer: The probability that a binomial random variable $X$ with parameters $n=6$ and $p=0.5$ is greater than 3 is \(\boxed{0.3438}\).