Math, asked by BrainlyHelper, 1 year ago

A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither an ace nor a king.

Answers

Answered by nikitasingh79
385
A deck of playing cards consists of 52 cards out of which 26 are black cards and  other 26 are red cards, where red cards consists of 13 cards of heart , 13 cards of diamond and black cards consists of 13 cards  of spades and 13 cards are club.
13 cards in each suit are king ,queen, Jack, 10, 9,8,7,6, 5, 4, 3 and 2.

King, queen ,and jack are called face cards. Total number of face cards are 12.

SOLUTION:
Total number of outcomes = 52

Favourable outcomes = four cards are of Ace & 4 cards are of king = 8
Total number of favourable outcomes = 52-8= 44

Probability = Number of favourable outcomes/ Total number of outcomes.

Required probability = P(neither an ace nor a king) = 44/52= 11/13.

Hence, the probability that the card  drawn is neither an ace nor a king 11/13.

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Anonymous: How can u answer by 1 min?
Answered by GauravSaxena01
75
Hey..!!!

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Total number of playing cards = 52
So the number of ways of drawing 1 card out of 52 cards =52
So n(S) = 52
(1) Drawn card is neither a heart nor a king
So probability of this can be obtained by

(1- probability that the drawn card is heart or a king)
So total number of hearts = 13
Total number of kings = 4
and 1 card is both heart and king 
So P(heart or king) =P(heart)+P(king)-P(heart and king) or P(heart or king)
=13/52+4/52−1/52=16/52
So P(neither heart nor king)=1-P
(heart or king)=1−16/52=36/52=9/13
(2)  Drawn card is neither an ace nor a king
There are 4 aces and 4 kings  ,  so total 8 cards

So P(neither ace nor king)=1-P(ace or king)=1−8/52=11/13

(3)Drawn card is neither a red nor a queen 

Number of red cards = 26
Number of queens =4

and  number of cards  which is queen and red =2
So P(drawn card is neither red nor queen)= 1-[P( red card)+P( queen)-P( red and queen)]
so P
(drawn card is neither red nor queen)
=1−(26/52+4/52−2/52)
=1−28/52=24/52=6/13

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I Hope this may be helpful for you......!!!!!
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