A card is drawn from a deck of 52 cards. Find the probability of getting an ace or a spade card.

Given: As a card is drawn from a deck of 52 cards.

Let ‘S’ denotes the event of card being a spade and ‘K’ denote the event of card being ace.

As we know that a deck of 52 cards contains 4 suits (Heart, Diamond, Spade and Club) each having 13 cards. The deck has 4 ace cards one from each suit.

We know that probability of an event E is given as-

By using the formula,

P (E) =\(\dfrac{ favourable \;outcomes }{ total \;possible \;outcomes}\)

= \( \dfrac{n (E) }{ n (S)}\)

Where, n (E) = numbers of elements in event set E

And n (S) = numbers of elements in sample space.

Hence,

P (S) =\(\dfrac{ n (spade) }{total\; number \;of\;cards}\)

= \( \dfrac{13 }{ 52}\)

= \( \dfrac{1}{4}\)

P (K) = \( \dfrac{4 }{ 52}\)

= \( \dfrac{1}{ 13}\)

And P (S ⋂ K) = \( \dfrac{1 }{ 52}\)

We need to find the probability of card being spade or ace, i.e.

P (Spade ‘or’ ace) = P(S ∪ K)

So, by definition of P (A or B) under axiomatic approach (also called addition theorem) we know that:

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

So, P (S ∪ K) = P (S) + P (K) – P (S ∩ K)

= \( \dfrac{1}{4}\) + \( \dfrac{1 }{ 13}\) – \( \dfrac{1 }{ 52}\)

= \( \dfrac{17 }{ 52}\) –\( \dfrac{1 }{ 52}\)

= \( \dfrac{16 }{ 52}\)

= \( \dfrac{4 }{ 13}\)

P (S ∪ K) = \( \dfrac{4 }{ 13}\)