Question

from a well shuffled pack of cards a card is drawn in random find the probability of getting a (1) black queen, (2) neither a red card or a queen

Answer 1

Hope this answer helps you!!!!

Answer 2

Hi there!
Here's the answer:

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¶¶¶ POINTS TO REMEMBER:

• Probability of Occurrence of Event E
$P(E) = \frac{No.\: of Favourable\: Outcomes\: for\: occurrence\: of\: E}{Total\: No.\: of outcomes}$

• In a pack of cards, there are a total of 52 cards.

• These 52 cards are categorized into 4 groups :

Spades, Clubs, Diamonds& Hearts.

• Out of the 52 cards,
26 cards are Black (Spades and Clubs) and 26 cards are Red(Diamonds and Hearts)


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¶¶¶ SOLUTION:

Let S be Sample Space
n(S)- No. of ways of drawing a card from 52 cards
=> n(S) = 52C1 = 52

(1) Let E1 be the Event that the card drawn is a black queen.

Possible cases for occurrence of Event E1:
• Queen of Spades
• Queen of Clubs
E1 = {Queen of Spades, Queen of Clubs}

n(E1) - No. of favourable cases for Occurrence of Event E1
(No. of elements in Set E1)

=> n(E1) = 2

Probability $P(E1) = \frac{n(E1)}{n(S)}$

$P(E1) = \frac{2}{52} = \frac{1}{26}$


(2) Let E2 be the Event that card drawn is neither a red card nor a queen

Possible cases for occurrence of Event E2:
• Except red cards (=> 26 black cards)
• Exclude 4 queens

E2 = {(All black cards} - {(Black queens)}
E2 = {(All black cards} - {Queen of Clubs, Queen of Spades}

n(E2) - No. of favourable cases for Occurrence of Event E2

=> n(E2) = 26 - 2 = 24

Probability $P(E2) = \frac{n(E2)}{n(S)}$

$P(E2) = \frac{24}{52} = \frac{6}{32}$

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