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 ' \(2^{\prime}\) 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 \(=2 / 52=1 / 26\).
