Question

Find the probability of drawing an Ace or a spade from a well shuffled pack of $52$ playing cards.

Answer 1

Hint : Probability of any given event is equal to the ratio of the favourable outcomes with
the total number of the outcomes. Probability is the state of being probable and the extent to which something is likely to happen in the particular situations.
$P(A) =$ $\dfrac{{Total{\text{ }}number{\text{ }}of{\text{ }}the{\text{ }}favourable{\text{ }}outcomes\;\;}}{{Total{\text{ }}number{\text{ }}of{\text{ }}the{\text{ }}outcomes}}$

Complete step-by-step answer :
Let A be an event of drawing an Ace or a Spade from the well shuffled pack of $52$ playing cards.
Total number of possible outcomes $= 52$
Therefore, $n(s) = 52$
Total number of an Ace cards in the pack of playing cards $= 4$ Cards
Total number of the spade cards in the pack of playing cards $= 13$ Cards
Total number of favourable outcomes
$=$ Total number of Ace cards $+$ total number of the spade cards $- 1$ [because $1$ ace card is spade]
$= 4 + 13 - 1$
$= 17 - 1 \\\ = 16 \\\$
Therefore, $n(A) = 16$
Now, Probability of the event of drawing an Ace or the spade is –
$\Rightarrow P(A) =$ Total number of the favourable outcomes / Total number of the outcomes
$\Rightarrow P(A) = \dfrac{{n(A)}}{{n(s)}}$
Put values,
$\Rightarrow P(A) = \dfrac{{16}}{{52}}$
$\Rightarrow P(A) = \dfrac{4}{{13}}$
Hence, the required solution is - The probability of drawing an Ace or a spade from a well shuffled pack of $52$ playing cards is $\dfrac{4}{{13}}$ .

Note : The probability of any event always lies between $0$ and $1$ . It can never be negative or the number greater than one. The impossible event is always zero. The probability of the events where event A or event B occur is known as the probability of the union of A and B. It is denoted by $P(A \cup B)$