Question

Find the probability of getting a king if a card is drawn from the pack of $52$ cards.
(a) $\dfrac{4}{{52}}$ or $\dfrac{1}{{13}}$
(b) $\dfrac{{13}}{{52}}$ or $\dfrac{1}{4}$
(c) $\dfrac{8}{{52}}$ or $\dfrac{2}{{13}}$
(d) Cannot be determined

Answer 1

The given problem revolves around the concepts of probability. So, we will use the definition of probability for each outcome or for the given outcome particularly. Using the formula for the probability i.e., if the probability of any event A is denoted by ‘P(A)’ and the formula to find the probability is $P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( s \right)}}$, where n(s) denotes the favorable outcomes and n(A) denotes the total number of outcomes for an respective event.

Complete step-by-step solution:
Since, the pack of $52$ cards is fair, all the thirteen outcomes, that is, ‘king’, ‘queen’, ‘jack’, ‘ace’, ‘two’, ‘three’, ‘four’, ‘five’, ‘six’, ‘seven’, ‘eight’, ‘nine’ and ‘ten’ (when drawn simultaneously or randomly) are equally probable;
We know that,
$Probability = \dfrac{\text{Number of outcomes}}{\text{Total outcomes in the event}} = \dfrac{{n\left( s \right)}}{{n\left( A \right)}}$ … ($1$)
Where, number of outcomes is, $n\left( s \right) = 4$ and,
Total outcomes in the event are, $n\left( A \right) = 52$ (in this case particularly)
As a result, there are total four possibilities of each thirteen outcomes (as discussed above) that is “Diamond”, “Heart”, “Club” and “Spade”
Mathematically, expressed as
$n\left( A \right) = 4 \times 13 = 52$ … (i)
It seems that, in this case the outcome “king” needs to be solved to get the possibilities out of the total outcomes that may be drawn in a pack of cards.
It may be,
One for Heart, one for Spade, one for Club, and one for Diamond
Hence, total outcomes for the event is,
$n\left( s \right) = 4$ … (ii)
From (i) and (ii),
Equation ($1$) becomes,
$\text{Probability} = \dfrac{4}{{52}}$
Or, considering its simplest form (i.e. the least multiple), that is dividing the equation by $4$, we get
$\text{Probability} = \dfrac{1}{{13}}$
$\Rightarrow$The option (a) is correct.

Note: One should know the basic formula of probability while solving a question. We should also know the concepts of independent events and that their probabilities are not affected by each other’s occurrences. We should take care of the calculations so as to be sure of our final answer. Also, we should know that the sum of probabilities of all the possible events in a situation is one.