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

(vi) spade or an ace

(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 |  51

1 Answer

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

Related Questions

One number is chosen from numbers 1 to 100. Find the probability that it is divisible by 4 or 6?

Class 11 Maths Probability View Answer

The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75. What is the probability of passing the Hindi examination?

Class 11 Maths Probability View Answer

A card is drawn from a deck of 52 cards. Find the probability of getting an ace or a spade card.

Class 11 Maths Probability View Answer

A die is thrown twice. What is the probability that at least one of the two throws come up with the number 3?

Class 11 Maths Probability View Answer

A natural number is chosen at random from amongst first 500. What is the probability that the number so chosen is divisible by 3 or 5?

Class 11 Maths Probability View Answer