3)) Given a deck of cards find the following probability, Put your answers in reduced fraction form or round them to the nearest ten-thousandy P (Drawing six oards (with replacement) and exactly 3 of the cards being black)

We have 6 draws with replacement, and we want exactly 3 of these cards to be black. Since the deck is evenly split between black and red cards, the probability of drawing a black card in one draw is \[ p = \frac{26}{52} = \frac{1}{2}. \] This is a binomial probability problem with parameters \(n = 6\) and \(p = \frac{1}{2}\). The probability of getting exactly \(k = 3\) black cards is given by the binomial formula \[ P(X=3) = \binom{6}{3} \left(\frac{1}{2}\right)^{3} \left(\frac{1}{2}\right)^{3} = \binom{6}{3} \left(\frac{1}{2}\right)^6. \] First, calculate the binomial coefficient: \[ \binom{6}{3} = \frac{6!}{3!\,3!} = \frac{720}{6 \times 6} = 20. \] Now, compute the probability: \[ P(X=3) = 20 \times \left(\frac{1}{2}\right)^6 = 20 \times \frac{1}{64} = \frac{20}{64} = \frac{5}{16}. \] Alternatively, as a decimal rounded to the nearest ten-thousandth: \[ \frac{5}{16} = 0.3125. \]