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)

Question

UpStudy AI · Accepted Answer

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:

$$
\frac{5}{16} = 0.3125.
$$
The probability of drawing exactly 3 black cards in 6 draws with replacement is $$\frac{5}{16}$$ or 0.3125.

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)

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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)

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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)