From a pack of cards, five of club and spade, jack of heart, ten of diamond and spade, and queen of club is put aside and then the remaining cards are well shuffled. Find the probability of getting

(a) a card of red queen

(b) a card of black queen

(c) a ten

(d) a five of club

(e) a jack

(f) an ace

(g) a face card

(h) a black face card

Total cards = 52
Cards that are kept aside are: five of club, five of spade, jack of heart(red), ten of diamond(red), ten of  spade, and queen of club (total 6 cards)

So number of cards left= 52-6=46
(a) a card of red queen

Since no red queens are kept aside, number of red queens = 2
Probability= 2/46=1/23

(b) a card of black queen
Since 1  black queen is kept aside, number of black queens = 1
Probability= 1/46

(c) a ten

Since two 10's are kept side, number of 10's = 2
Probability= 2/46=1/23

(d) a five of club

Since 5 of club is kept aside, number of 5's of club=0
Probability=0/46=0

(e) a jack
Since a jack is kept aside, number of Jack's left= 3
Probability=3/46

(f) an ace
Since no ace is kept aside, number of Aces = 4
Probability=4/46=2/23

(g) a face card
Face cards are King, Queen Jack.
So total number of face cards =12
Since 1 Queen and 1 Jack are kept aside, number of face cards = 12-2=10
Probability=10/46=5/23

(h) a black face card
Total number of black face cards = 2 Jacks+ 2 queens + 2 kings=6 cards
Since 1 queen is kept aside, number of face cards = 5
Probability= 5/46






 

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total cards=52

remaining cards=46

(a) 2/46=1/23

(b) 1/46

(c) 2/46=1/23

(d) 0

(e) 3/46

(f) 4/46= 2/23

(g) 10/46= 5/23

(h) 5/46

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how can we get remaining cards as 46 explain

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