Assume that a procedure yields a binomial distribution with $n=7$ trials and a probability of success of $p=0.50$. Use a binomial probability table to find the probability that the number of successes $\mathrm{x}$ is exactly 3 .
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$P(3)=\square$ (Round to three decimal places as needed.)

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Step 1 ：The probability of exactly x successes (k successes) in n trials is given by the formula for the binomial probability: \(P(x) = C(n, x) * (p^x) * ((1-p)^(n-x))\)

Step 2 ：In this case, n=7, p=0.50, and x=3. So, we can substitute these values into the formula: \(P(3) = C(7, 3) * (0.50^3) * ((1-0.50)^(7-3))\)

Step 3 ：First, calculate \(C(7, 3)\). This is the number of ways to choose 3 successes from 7 trials. \(C(7, 3) = \frac{7!}{3!(7-3)!} = \frac{7*6*5}{3*2*1} = 35\)

Step 4 ：Next, calculate \((0.50^3) = 0.125\) and \((1-0.50)^(7-3) = (0.50^4) = 0.0625\)

Step 5 ：Finally, substitute these values back into the formula: \(P(3) = 35 * 0.125 * 0.0625 = 0.273\)

Step 6 ：So, the probability that the number of successes x is exactly 3 is \(\boxed{0.273}\) (rounded to three decimal places).