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

The event in question [tex](\text{black suit} \; \land\; \text{face card})[/tex] 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 [tex](\text{black suit} \; \land\; \text{face card})[/tex] into two smaller events that are easier to handle: event [tex]\rm A[/tex] and event [tex]\rm B[/tex]. The choice of event [tex]\rm A[/tex] and event [tex]\rm B[/tex] should ensure that [tex]\mathrm{A \; \lor\; B} = (\text{black suit} \; \land\; \text{face card})[/tex].

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

[tex]\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}[/tex].

One option involves

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

In other words:

[tex]\mathrm{A} = (\text{black suit})[/tex].
[tex]\mathrm{B} = (\text{face card}) \; \land (\lnot (\text{black suit}))[/tex].

Verify that:

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

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

[tex]\displaystyle P(\mathrm{A}) = \frac{26}{52}[/tex].

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

[tex]\displaystyle P(\mathrm{B}) = \frac{6}{52}[/tex].

Since event [tex]\rm A[/tex] and [tex]\rm B[/tex] are mutually-exclusive:

[tex]\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}[/tex].

Therefore:

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