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

there is a 26 out of 52 of a chance

Explanation:

Answer 2

`\displaystyle (8)/(13)`, which is approximately
`0.62`.

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 * 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}) = (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* 3 = 6` face cards that are not from a black suit. In other words, event

`\displaystyle P(\mathrm{B}) = (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}) \\ &= (26)/(52) + (6)/(52) = (8)/(13)\end{aligned}`.

Therefore:

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