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Re: One card is drawn at random from a pack of 52 cards. What is the proba [#permalink]
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D33T wrote:
Can someone help review my working method and help identify what I'm doing wrongly? I used the following approach but couldn't reach any of the answer choices.

In any given deck of 52 cards, there are:
Red = 26 cards --> Prob: 26/ 52 (i.e. 1/2)
'10' = 4 cards --> Prob: 4/52
Spades = 13 cards --> Prob: 13/52 (i.e. 1/4)

Probabilities of picking a single card with the following characteristics are:
Black (i.e. not red) = 1/2
'1-9, JQK' (i.e. not '10') = 48/52
Hearts, Clubs, Diamonds (i.e. not Spades) = 39/52 = 3/4

Therefore:
Prob (not red AND not '10' AND not spade) = 1/2 * 48/52 * 3/4 = 9/26


When multiplying probabilities you are saying both events are occurring.

Once you've established black in your first statement, red should be eliminated in your third. The third statement should read "clubs" 1/2, indicating a further partitioning of black from the first statement

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Re: One card is drawn at random from a pack of 52 cards. What is the proba [#permalink]
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Thanks so much! Regor60
Applying that approach, the probabilities work out to be -

Black (only clubs or spades): 26/52 (ie 1/2)
Not 10 (1-9, JQK): (26-2)/26 (ie 12/13)
Not Spade: 12/24 (ie 1/2)

Which gives:
1/2 * 12/13* 1/2 = 3/13 😃

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Re: One card is drawn at random from a pack of 52 cards. What is the proba [#permalink]
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KANNUPRIYA17 wrote:
One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is NEITHER a 'spade' NOR a'10' NOR a'Red'.


A.9/52

B.3/13

C.10/13

D.21/26

E.43/52



There are in total \(52 \) cards.

\(13\) Spades ( Black )
\(13\) Clubs (Black )
\(13 \) Diamonds (Red)
\(13 \) Hearts (Red)

We DON'T NEED :
\(13\) Spades ( Black )
\(13 \) Diamonds (Red)
\(13 \) Hearts (Red)
\(10 \) of Hearts, \(10\) of Diamonds , \(10 \) of Spades and \(10 \) of Clubs.

However we have already removed \(10 \) of Spades, \(10\) of Diamonds , \(10 \) of Hearts above when we removed \(13\) of each kind.

Hence we have yet to remove ONLY \(10\) of Clubs.

Hence total removals = \(13 +13 +13 +1 = 40\)

Left with \(52 - 40 = 12 \)

Probability = \(\frac{12}{52} = \frac{3}{13}\)

Ans=B

Hope it's clear.
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Re: One card is drawn at random from a pack of 52 cards. What is the proba [#permalink]
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Re: One card is drawn at random from a pack of 52 cards. What is the proba [#permalink]
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