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Andrews [41]
3 years ago
10

A card is drawn at random from a standard deck of cards. What is the probability that the card is black or a face card

Mathematics
2 answers:
LenaWriter [7]3 years ago
6 0

Answer:

there is a 26 out of 52 of a chance

Step-by-step explanation:

malfutka [58]3 years ago
5 0

Answer:

\displaystyle \frac{8}{13}, which is approximately 0.62.

Step-by-step explanation:

A standard playing card deck includes:

  • Twelve face cards (three for each suit.)
  • Two black suits.

The event in question (\text{black suit} \; \land\; \text{face card}) includes a large number of outcomes; it is a compound event. This event includes a large number of outcomes. One way to keep calculations simple is to split this event (\text{black suit} \; \land\; \text{face card}) into two smaller events that are easier to handle: event \rm A and event \rm B. The choice of event \rm A and event \rm B should ensure that \mathrm{A \; \lor\; B} = (\text{black suit} \; \land\; \text{face card}).

Note that \rm A and \rm B should be mutually exclusive (i.e., P(\mathrm{A \; \land \; B}) = 0) to ensure that:

\begin{aligned}&P(\text{black suit} \; \land\; \text{face card}) \\ &= P(\mathrm{A \lor B})\\ &= P(\mathrm{A}) + P(\mathrm{B}) - P(\mathrm{A \land B}) \\ &= P(\mathrm{A}) + P(\mathrm{B}) \end{aligned}.

One option involves

  • letting \rm A be the event that the card is from a black suit, and
  • letting \rm B be the event that the card is a face card and is not from a black suit.

In other words:

  • \mathrm{A} = (\text{black suit}).
  • \mathrm{B} = (\text{face card}) \; \land (\lnot (\text{black suit})).

Verify that:

  • \rm A and \rm B are mutually exclusive, and that
  • \rm A \; \lor \; B is the same as (\text{black suit} \; \land\; \text{face card}).

Note that event \rm A is itself a compound event with 2 \times 13 = 26 possible outcomes, one for each card in the two black suits. Overall, the event space includes 52 outcomes (one for each card.) Since these outcomes are equally likely:

\displaystyle P(\mathrm{A}) = \frac{26}{52}.

Event \rm B is also a compound event. There are two red suits in a standard deck. Each suit includes three face cards. That corresponds to 2\times 3 = 6 face cards that are not from a black suit. In other words, event

\displaystyle P(\mathrm{B}) = \frac{6}{52}.

Since event \rm A and \rm B are mutually-exclusive:

\begin{aligned} & P(\mathrm{A} \; \lor \; \mathrm{B}) \\&= P(\mathrm{A}) + P(\mathrm{B}) \\ &= \frac{26}{52} + \frac{6}{52} = \frac{8}{13}\end{aligned}.

Therefore:

\displaystyle P(\text{black suit} \; \land\; \text{face card}) = P(\mathrm{A}) + P(\mathrm{B}) = \frac{8}{13} \approx 0.62.

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Step-by-step explanation:

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This is what i get

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