Question

A card is drawn from a standard deck of 52 playing cards. What is the probability that the card is an ace or a king?

Answer 1

Hint: We will use the concepts of probability of events to solve this problem. And also, we will know about standard deck of cards and their constitution and properties. We will also use some formulas related to combinations to solve this problem and which makes the problem easier.

Complete step-by-step solution:
Generally, a standard deck of cards has 52 cards in it.
Out of which, 13 cards are spades, 13 cards are hearts, 13 cards are diamonds and the other 13 cards are clubs.
And, all these 4 sets i.e., clubs, hearts, spades and diamonds have cards with numbers \[2 - 10\] along with a king, queen, joker and an ace.
So, finally, every deck of cards has 4 aces, 4 kings, 4 queens, and 4 jokers.
 In mathematics, probability is defined as the occurrence of a random event. It is also defined as the ratio of number of favorable outcomes to total number of outcomes. The probability of any event always lies in the range \[[0,1]\] .
So, \[P(E) = \dfrac{{{\text{number of favorable outcomes}}}}{{{\text{total number of outcomes}}{\text{.}}}}\] is the probability of an event E.
If the probability of an event is 0, then the event doesn’t happen.
If the probability of an event is 1, then it will happen for sure.
So, now, in the question, it is given that a card is taken from a deck of 52 cards.
Let the event of getting an ace or a king be E.
So, the total number of outcomes is \[{}^{52}{C_1}\] which is equal to 52. (As we need to select one card from 52)
And the number of ways of getting an ace from four aces is \[{}^4{C_1}\] which is equal to 4.
And the number of ways of getting a king from four kings is \[{}^4{C_1}\] which is also equal to 4.
So, total number of favourable outcomes is equal to \[4 + 4 = 8\]
So, probability of getting an ace or a king is equal to
\[P(E) = \dfrac{8}{{52}} = \dfrac{2}{{13}}\]

Note: The number of ways of selecting \[n\] things out of \[m\] is equal to \[{}^m{C_n}\] . And its value is \[{}^m{C_n} = \dfrac{{m!}}{{n!\left( {m - n} \right)!}}\] . And also make a note that, probability of an event will never be a negative value or a value greater than 1. If you get such value, then you have committed a mistake in your solution.