Question

From a pack of 52 cards, 2 cards are drawn at random. What is the probability that either both are black or both are kings?

Maths : Probability​

Answer 1

ANSWER:

Given:

• 2 cards are drawn at random from a pack of 52 cards

To Find:

• Probability that either both are black or both are kings

Solution:

Let the Sample S = Drawing either 2 black cards or 2 king cards.

Let A be the set of possibilities in which 2 black cards can be drawn, and B be the set of possibilities in which 2 king cards can be drawn.

So,

⇒ A ∪ B be the set of all possibilities in which either 2 black cards or 2 king cards can be drawn.

Also, we know that, there are 2 king cards which are black in color.

That is,

⇒ A ∩ B be the set of possibilities in which the cards drawn are 2 black kings.

Now, the number of total possibilities are,

$\implies\sf n(A ∪ B)=^{52}C_2$

The number of possibilities in which the cards drawn are 2 black kings are,

$\implies\sf n(A ∩ B)=^{2}C_2$

The number of possibilities of drawing 2 black cards are,

$\implies\sf n(A)=^{26}C_2$

(There are 26 black cards)

And,

The number of possibilities of drawing 2 king cards are,

$\implies\sf n(A)=^{4}C_2$

(There are 4 king cards)

We know that,

$\hookrightarrow\sf n(X ∪ Y) = n(X) + n(Y) - n(X ∩ Y)$

So,

$\implies\sf n(A ∪ B) = n(A) + n(B) - n(A ∩ B)$

The required probability will be,

$\implies\sf P(S) = \dfrac{n(A) + n(B) - n(A ∩ B)}{n(A ∪ B)}$

Substituting the values,

$\implies\sf P(S) = \dfrac{^{26}C_2+^4C_2-^2C_2}{^{52}C_2}$

$\implies\sf P(S) = \dfrac{(26\times25)+(4\times3)-(2\times1)}{52\times51}$

$\implies\sf P(S) = \dfrac{650+12-2}{52\times51}$

$\implies\sf P(S) = \dfrac{660}{52\times51}$

Dividing the numerator and denominator by 12,

$\implies\bf P(S) = \dfrac{55}{221}$

Therefore, the probability of drawing 2 cards at random which are either both black or both kings is 55/221.