Question

A card is drawn at random well shuffled packed of 52 cards. Find the probability of getting

i ) a face card

ii ) the queen of diamonds

iii ) a card of spade or an ace

iv ) neither a jack nor a king.

Answer 1

$\huge \bf \pink{Hey \: there !! }$


→ Here is some facts about playing cards .

1 . A deck of playing cards has 52 cards .

2. There are 4 suits, namely (i)Spades , (ii) Clubs , (iii) Hearts , and (iv) Diamonds . There are 13 cards of each suits .

3. Spades and clubs are black cards . ( 26 cards . )

4. Hearts and diamonds are red cards . ( 26 cards . )

5. There 12 face cards , namely 4 kings, 4 queens and 4 jacks .


▶ Now, A/Q

(i) A face card .

→ Total number of face card = 12 .

•°• P(E) of face card = ( Number of face card )/( Total number of cards ) .

= 12/52 .

= 3/13 .


(ii) A queen of diamond .

→ Total number of queen of diamond = 1 .

•°• P(E) of queen of diamond = ( Number of queen of diamond)/( Total number of cards ) .

= 1/52 .


(iii) A card of spades or an ace .

→ There are 13 cards of spades including one ace and there are 3 more aces .

→ So, the number of favourable cases = 13 + 3 = 16 .

•°• P(E) of getting a card of spades or an ace

= 16/52 .

= 4/13 .


(iv) Neither a jack nor a king .

→ There are 4 jacks and 4 kings .

→ So, the number of cards which are neither a jack nor a kings = [ 52 - ( 4 + 4 ) ] = 44 .

•°• P(E) of getting a card neither a jack nor a king

= 44/52 .

= 11/13 .


✔✔ Hence, it is solved ✅✅.



THANKS


#BeBrainly.

Answer 2

$\huge \bf{hey}$

$\huge{ \mathfrak{ \red{here \: is \: answer}}}$

$\bf{given}$
total number of card= 52

number of face card= 12

number of queen of diamond=1

number of spade=13

number of ace= 4

13+3=16

number of card except jack or king=52-4+4
52-8=44

1. probability of obtaining face card= 12/52

3/13

2. probability of obtaining queen of diamond= 1/52



3. probability of spade or an ace= 16/52

4/13

4. probablity of niether jack nor king=

44/52= 11/13

$\huge \boxed{ \boxed{ \boxed{ \blue{hope \: it \: helps}}}}$

$\huge{ \mathfrak{thanks}}$