One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is:

0 votes
15 views
asked Feb 9, 2018 in Mathematics by Kundan kumar (49,132 points) 34 379 1010

One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is:
(i) an ace. (ii) red. (iii) either red or king. (iv) red and a king. (v) a face card. (vi) a red face card. (vii) '2' of spades. (viii) '10' of a black suit.

1 Answer

0 votes
answered Feb 9, 2018 by Vikash Kumar (144,729 points) 8 11 21
 
Best answer

Out of 52 cards, one card can be drawn in 52 ways.
So, total number of elementary events = 52

(i) There are four ace cards in a pack of 52 cards. So, one ace can be chosen in 4 ways.

Favorable number of elementary events = 4

Hence, required probability =4/52=1/13

(ii) There are 26 red cards in a pack of 52 cards. Out of 26 red cards, one card can be chosen in 26 ways.

Favourable number of elementary events= 26

Hence, required probability=26/52=1/2

(iii) There are 26 red cards, including two red kings, in a pack of 52 playing cards. Also, there are 4 kings, two red and two black. Therefore, card drawn will be a red card or a king if it is any one of 28 cards (26 red cards and 2 black kings).

Favourable number of elementary events= 28

Hence, required probability=28/52=7/13.

(iv) A card is drawn will be red as well as king, if it is a  red king. There are 2 red kings in a pack of 52 playing cards.

Favourable number of elementary events= 2

Hence, required probability=2/52=1/26.

(v) In a deck of 52 cards: kings, queens, and jacks are called face cards. Thus, there are 12 face cards. So, one face card can be chosen in 12 way.

Favourable number of elementary events=12

Hence, required probability=12/52=3/13.

(vi) There are 6 red face cards 3 each from diamonds and hearts. Out of these 6 red face cards, one card can be chosen in 6 ways.

Favourable number of elementary events= 6

Hence, required probability=6/52=3/26.

(vii) There is only one '2' of spades.

Favourable number of elementary events= 1

Hence, required probability=1/52.

(viii) There are two suits of black cards viz. spades and clubs. Each suit contains one card bearing number 10.

Favourable number of elementary events= 2

Hence, required probability=2/52=1/26.

...