A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of its being a spade or a king.

Asked by Aaryan | 1 year ago |  41

##### Solution :-

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 King.

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 king 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 king, i.e.

P (Spade ‘or’ King) = 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) = $$\dfrac{4 }{ 13}$$

Answered by Aaryan | 1 year ago

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