A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is

Given: Pack of 52 cards.

We know that, a card is drawn from a pack of 52 cards, so number of elementary events in the sample space is

n (S) = ⁵²C₁ = 52

(i) Let E be the event of drawing a black king

n (E) =²C₁ =2 (there are two black kings one of spade and other of club)

P (E) = \( \dfrac{ n (E) }{ n (S)}\)

= \(\dfrac{ 2 }{ 52}\)

= \( \dfrac{ 1 }{26}\)

(ii) Let E be the event of drawing a black card or a king

n (E) = ²⁶C₁+⁴C₁–²C₁= 28

[We are subtracting 2 from total because there are two black king which are already counted and to avoid the error of considering it twice.]

P (E) = \( \dfrac{ n (E) }{ n (S)}\)

=\( \dfrac{ 28 }{52}\)

=\( \dfrac{ 7 }{13}\)

(iii) Let E be the event of drawing a black card and a king

n (E) =²C₁ = 2 (there are two black kings one of spade and other of club)

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

=\( \dfrac{ 2 }{52}\)

= \( \dfrac{ 1 }{26}\)

(iv) Let E be the event of drawing a jack, queen or king

n (E) = ⁴C₁+⁴C₁+⁴C₁ = 12

P (E) = \( \dfrac{ n (E) }{ n (S)}\)

= \(\dfrac{ 12 }{ 52}\)

= \( \dfrac{ 3 }{ 13}\)

(v) Let E be the event of drawing neither a heart nor a king

Now let us consider E′ as the event that either a heart or king appears

n (E′) = ⁶C₁+⁴C₁-1=16 (there is a heart king so it is deducted)

P (E′) = \( \dfrac{ n (E') }{ n (S)}\)

= \( \dfrac{ 16 }{ 52}\)

= \( \dfrac{4 }{ 13}\)

So, P (E) = 1 – P (E′)

= 1 – \( \dfrac{4 }{ 13}\)

= \( \dfrac{9 }{ 13}\)

(vi) Let E be the event of drawing a spade or king

n (E)=¹³C₁+⁴C₁-1=16

P (E) = \( \dfrac{ n (E) }{ n (S)}\)

= \( \dfrac{16}{ 52}\)

= \( \dfrac{4 }{ 13}\)

(vii) Let E be the event of drawing neither an ace nor a king

Now let us consider E′ as the event that either an ace or king appears

n(E′) = ⁴C₁+⁴C₁ = 8

P (E′) =\( \dfrac{ n (E') }{ n (S)}\)

= \( \dfrac{8 }{ 52}\)

= \( \dfrac{2 }{ 13}\)

So, P (E) = 1 – P (E′)

= 1 – \( \dfrac{2 }{ 13}\)

= \( \dfrac{11 }{ 13}\)

(viii) Let E be the event of drawing a diamond card

n (E)=¹³C₁=13

P (E) = \( \dfrac{ n (E) }{ n (S)}\)

= \( \dfrac{13 }{ 52}\)

= \( \dfrac{1}{4}\)

(ix) Let E be the event of drawing not a diamond card

Now let us consider E′ as the event that diamond card appears

n (E′) =¹³C₁=13

P (E′) =\( \dfrac{ n (E') }{ n (S)}\)

= \( \dfrac{13 }{ 52}\)

= \( \dfrac{1 }{ 4}\)

So, P (E) = 1 – P (E′)

= 1 – \( \dfrac{1 }{ 4}\)

= \( \dfrac{3}{4}\)

(x) Let E be the event of drawing a black card

n (E) =²⁶C₁ = 26 (spades and clubs)

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

=\( \dfrac{26 }{ 52}\)

= \( \dfrac{1}{2}\)

(xi) Let E be the event of drawing not an ace

Now let us consider E′ as the event that ace card appears

n (E′) = ⁴C₁ = 4

P (E′) = \( \dfrac{ n (E') }{ n (S)}\)

= \( \dfrac{4}{ 52}\)

= \( \dfrac{1 }{ 13}\)

So, P (E) = 1 – P (E′)

= 1 –\( \dfrac{1 }{ 13}\)

=\( \dfrac{12 }{ 13}\)

(xii) Let E be the event of not drawing a black card

n (E) = ²⁶C₁ = 26 (red cards of hearts and diamonds)

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

= \( \dfrac{26 }{52}\)

= \( \dfrac{1}{2}\)

Answered by Aaryan | 1 year ago