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Alecsey [184]
3 years ago
8

Two cards are selected from a standard deck of 52 playing cards. The first card is not replaced before the second card is select

ed . Find the probability of selecting a black card and selecting a red card. The probability of selecting a black card and then selecting a red card is
Mathematics
1 answer:
Blizzard [7]3 years ago
5 0

Answer:

Probability is:   $ \frac{\textbf{13}}{\textbf{51}} $

Step-by-step explanation:

From a deck of 52 cards there are 26 black cards. (Spades and Clubs).

Also, there are 26 red cards. (Hearts and Diamonds).

First, we determine the probability of drawing a black card.

P(drawing a black card) = $ \frac{number \hspace{1mm} of  \hspace{1mm} black  \hspace{1mm} cards}{total  \hspace{1mm} number  \hspace{1mm} of  \hspace{1mm} cards} $  $ = \frac{26}{52} = \frac{\textbf{1}}{\textbf{2}} $

Now, since we don't replace the drawn card, there are only 51 cards.

But the number of red cards is still 26,

∴ P(drawing a red card) = $ \frac{number  \hspace{1mm} of  \hspace{1mm} red  \hspace{1mm} cards}{total  \hspace{1mm} number  \hspace{1mm}of  \hspace{1mm} cards} $  $ = \frac{26}{51}  $

Now, the probability of both black and red card = $ \frac{1}{2} \times \frac{26}{51} $

$ = \frac{\textbf{13}}{\textbf{51}} $

Hence, the answer.

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\huge\text{Hey there!}

\mathsf{13x = 6}

\huge\textsf{DIVIDE 13 to BOTH SIDES}

\mathsf{\dfrac{13x}{13}=\dfrac{6}{13}}

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\rm{KEEP: \dfrac{6}{13} \ because \ that \ gives \ you \ the \ value \ of \ x}

\boxed{\boxed{\mathsf {Answer: x = \bf \dfrac{6}{13}}}}\huge\checkmark

\boxed{\textsf{or you could say \boxed{\mathsf{x = \underline{\bf 0.461538}}} they both equal to the same thing}}\huge\checkmark

\large\text{Good luck on your assignment and enjoy your day!}

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