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

(i) a black king

(ii) either a black card or a king

(iii) black and a king

(iv) a jack, queen or a king

(v) neither an ace nor a king

(vii) neither an ace nor a king

(viii) a diamond card

(ix) not a diamond card

(x) a black card

(xi) not an ace

(xii) not a black card

Asked by Aaryan | 1 year ago |  31

##### Solution :-

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+4C12C1= 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+4C= 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+4C= 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) =26C= 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′) = 4C= 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) = 26C= 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

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