Question

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

Answer 1

Answer:

there is a 26 out of 52 of a chance

Step-by-step explanation:

Answer 2

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:

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