Math, asked by selfiequeen979, 7 months ago

One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:

(i) a king of red colour

(ii) a face card

(iii) a red face card

(iv) the jack of hearts

(v) a spade

(vi) the queen of diamonds.

Answers

Answered by thanushiya72
8

Step-by-step explanation:

In 52 cards getting a one card as 52 times

(i)The king in card is two getting in 2 times

Then probability = 2/52 = 1/26

(ii)In 52 cards the king ,queen and joker are three face card.Then total face cards are 12.So these card get 12 times

Then probability = 12/52 =3/13

(!!!)In 52 cards the red face card are 3 face card.Then total

red face cards are 3.So these card get 6 times

Then probability = 6/52 =3/26

(iv)In 52 cards the jack of heart card is one .So the getting time card is one

Then probability =1/52

(v)In 52 cards the spade cards are 13. So getting time is 13

Then probability = 13/52 =1/4

(vi)In 52 cards the queen of diamonds is one .So getting time is one

Then probability= 1/52

Answered by Anonymous
40

Given:

  • 1 Card is Drawn from a well-shuffled deck of 52 Cards

Find:

What is the Probability of :-

  • A king of Red Color
  • A face Card
  • A red Face Card
  • The jack of Hearts
  • A spade
  • The queen of diamonds

Solution:

A king of Red Color:-

Total Possible Outcomes = 52

we, know

There are two suits of Red Cards, i.e., diamond \red\diamondsuit and heart \red\heartsuit. Each suit contains one king.

\therefore Favourable Outcomes = 1 + 1 = 2

we, know that

 \boxed{\rm P(E_{1}) =  \dfrac{Favourable \: Outcomes}{Total \: Outcomes}}

where,

  • Favourable Outcomes = 2
  • Total Outcomes = 52

So,

 \red\multimap\rm P(E_{1}) =  \dfrac{Favourable \: Outcomes}{Total \: Outcomes} \\  \\

 \green\multimap\rm P(E_{1}) =  \dfrac{2}{52} \\  \\

 \pink\multimap\rm P(E_{1}) =  \dfrac{1}{26} \\  \\

Hence, Probability of Getting a king of Red Color = \dfrac{1}{26}

 \rule{400}{1.5}

A face Card:-

Total Possible Outcomes = 52

we, know

There, are 12 face Card in a pack.

\therefore Favourable Outcomes = 12

we, know that

 \boxed{\rm P(E_{2}) =  \dfrac{Favourable \: Outcomes}{Total \: Outcomes}}

where,

  • Favourable Outcomes = 12
  • Total Outcomes = 52

So,

 \pink\multimap\rm P(E_{2}) =  \dfrac{Favourable \: Outcomes}{Total \: Outcomes} \\  \\

 \pink\multimap\rm P(E_{2}) =  \dfrac{12}{52} \\  \\

 \pink\multimap\rm P(E_{2}) =  \dfrac{3}{13} \\  \\

Hence, Probability of getting an Face Card = \dfrac{3}{13}

 \rule{400}{1.5}

A Red Face Card:-

Total Outcomes = 52

we, know

There are two suits of Red Cards, i.e., diamond \red\diamondsuit and heart \red\heartsuit. Each suit contains 3 face Cards.

\therefore Favourable Outcomes = 2×3 = 6

we, know that

 \boxed{\rm P(E_{3}) =  \dfrac{Favourable \: Outcomes}{Total \: Outcomes}}

where,

  • Favourable Outcomes= 6
  • Total Outcomes = 52

So,

 \orange\multimap\rm P(E_{3}) =  \dfrac{Favourable \: Outcomes}{Total \: Outcomes} \\  \\

 \orange\multimap\rm P(E_{3}) =  \dfrac{6}{52} \\  \\

 \orange\multimap\rm P(E_{3}) =  \dfrac{3}{26} \\  \\

Hence, Probability of Getting a Red Face Card = \dfrac{3}{26}

 \rule{400}{1.5}

The Jack of Hearts

Total Outcomes = 52

we, know

There are only one jack of heart \heartsuit

\therefore Favourable Outcomes = 1

we, know that

 \boxed{\rm P(E_{4}) =  \dfrac{Favourable \: Outcomes}{Total \: Outcomes}}

where,

  • Favourable Outcomes = 1
  • Total Outcomes = 52

So,

 \blue\multimap\rm P(E_{4}) =  \dfrac{Favourable \: Outcomes}{Total \: Outcomes} \\  \\

 \blue\multimap\rm P(E_{4}) =  \dfrac{1}{52} \\  \\

Hence, Probability of Getting a jack of Hearts \heartsuit = \dfrac{1}{52}

 \rule{400}{1.5}

The Spade:-

Total Outcomes = 52

There are 13 Cards of Spades \spadesuit

\therefore Favourable Outcomes = 13

we, know that

 \boxed{\rm P(E_{5}) =  \dfrac{Favourable \: Outcomes}{Total \: Outcomes}}

where,

  • Favourable Outcomes = 13
  • Total Outcomes = 52

So,

 \purple\multimap\rm P(E_{5}) =  \dfrac{Favourable \: Outcomes}{Total \: Outcomes} \\  \\

 \purple\multimap\rm P(E_{5}) =  \dfrac{13}{52} \\  \\

 \purple\multimap\rm P(E_{5}) =  \dfrac{1}{4} \\  \\

Hence, Probability of getting a Spade = \dfrac{1}{4}

 \rule{400}{1.5}

The queen of Diamonds:-

Total Outcomes = 52

we, know

There is only 1 queen of Diamonds \diamondsuit

\therefore Favourable Outcomes = 1

we, know that

 \boxed{\rm P(E_{6}) =  \dfrac{Favourable \: Outcomes}{Total \: Outcomes}}

where,

  • Favourable Outcomes = 1
  • Total Outcomes = 52

So,

 \green\multimap\rm P(E_{6}) =  \dfrac{Favourable \: Outcomes}{Total \: Outcomes} \\  \\

 \green\multimap\rm P(E_{6}) =  \dfrac{1}{52} \\  \\

Hence, Probability of Getting a queen of Diamond \diamondsuit = \dfrac{1}{52}


Anonymous: Awesome!
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