What is the probability that a card drawn at random from a pack of 52 cards is either a king or spade?

If A and B denote the events of drawing a spade card and a king, respectively, then event A consists of 13 sample points, whereas event B consists of four sample points.
Thus, 

\[P\left( A \right) = \frac{13}{52}\] and

\[P\left( B \right) = \frac{4}{52}\] 

The compound event (A ∩ B) consists of only one sample point, i.e. the king of spade.
So,

\[P\left( A \cap B \right) = \frac{1}{52}\]

By addition theorem, we have:
P (A ∪ B) = P(A) + P (B) -P (A ∩ B)
                = \[\frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{13 + 4 - 1}{52} = \frac{16}{52} = \frac{4}{13}\]

Hence, the probability that the card drawn is either a spade or a king is given by \[\frac{4}{13} .\]

Text Solution

Solution : Let S be the sample space. Then, n(S) = 52. <br> Let `E_(1)` = event of getting a spade, <br> and `E_(2)` = event of getting a king. <br> Then, `(E_(1) nn E_(2)) =` event of getting a king of spades. <br> Clearly, `n(E_(1)) = 13, n(E_(2)) = 4 and n(E_(1) nn E_(2)) = 1.` <br> `therefore P(E_(1)) = (n(E_(1)))/(n(S)) = 13/52 = 1/4, P(E_(2)) = (n(E_(2)))/(n(S)) = 4/52 = 1/13.` <br> and `P(E_(1) nn E_(2)) = (n(E_(1) nn E_(2)))/(n(S)) = 1/52.` <br> `therefore` P(getting a spade or a king) <br> `= P(E_(1) or E_(2)) = P(E_(1) uu E_(2))` <br> `= P(E_(1)) + P(E_(2)) - P(E_(1) nn E_(2))` [by the addition theorem for two events] <br> `= (1/4 + 1/13 - 1/52) = 16/52 = 4/13.` <br> Hence, the required probability is `4/13`.

Correct Answer:

Description for Correct answer:

Required probability = \( \large\frac{3}{52} + \frac{13}{52} = \frac{16}{52} = \frac{4}{13} \)[Hint 13 / 52 because there are 13 spades and 3 / 52 instead of 4 / 52 (there are four kings) because one king is already counted in spades.]

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