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Bunuel
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If two cards are picked out of a standard 52-card deck of cards, witho 29 Jul 2022, 02:49
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If two cards are picked out of a standard 52-card deck of cards, without replacement, what is the probability that the cards picked are of two different suits ? (A standard 52-card deck of cards has four suites: hearts, clubs, spades, diamonds (13 cards in each suit).)

A. 15/17
B. 13/17
C. 11/17
D. 4/17
E. 13/102

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pintukr
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If two cards are picked out of a standard 52-card deck of cards, witho Updated on: 16 Aug 2022, 10:33
Originally posted by pintukr on 29 Jul 2022, 06:38.
Last edited by pintukr on 16 Aug 2022, 10:33, edited 1 time in total.
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If two cards are picked out of a standard 52-card deck of cards, without replacement, what is the probability that the cards picked are of two different suits ? (A standard 52-card deck of cards has four suites: hearts, clubs, spades, diamonds (13 cards in each suit).)

A. 15/17
B. 13/17
C. 11/17
D. 4/17
E. 13/102

There are 4 cases possible

ie, First pick can be either a Heart / Club / Spade / Diamond (all equally probable)

(say) Probabilty (pick a Heart) = 13/52 = 1/4

Without replacement,
Probability (pick a Card, which is NOT a Heart) = 39/51 = 13/17
(39 cards ie, 13*3 belong to Club / Spade / Diamond; while Total cards remaining =51)

Hence, Total probability (when Heart is chosen as the first pick) = (1/4) * (13/17)

Similarly,
Total probability (when Spade is chosen as the first pick) = (1/4) * (13/17)
Total probability (when Club is chosen as the first pick) = (1/4) * (13/17)
Total probability (when Diamond is chosen as the first pick) = (1/4) * (13/17)

Since all the aforementioned events are mutually exclusive, Overall probability = 4 * (1/4) * (13/17)
= 13/17

Correct option is (B)
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If two cards are picked out of a standard 52-card deck of cards, witho 29 Jul 2022, 08:19
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Pick a card. That establishes the suit.

There are now 12 cards remaining of that suit out of 51 remaining cards, leaving 39 cards representing different suits.

Picking any of these 39 cards will satisfy the question.

39/51=13/17

Posted from my mobile device
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If two cards are picked out of a standard 52-card deck of cards, witho 05 Oct 2024, 10:49
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we can pick one card in 52 ways, the card of the same suit in 12 ways (without replacement). we can pick two card without replacement in 52*51 ways

thus the probability of getting the same suit is

(52*12)/(52/51) = 4/17
thus the probability of getting different suits is 1-4/17 = 13/17
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If two cards are picked out of a standard 52-card deck of cards, witho 07 Oct 2024, 04:36
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Random probability is 13/52*39/51, multiply this by 4 to get the possibilities of all the 4 suits. Ans 13/17
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If two cards are picked out of a standard 52-card deck of cards, witho 25 Oct 2024, 14:04
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Probability of not getting the same card =52*39/52*51=39/51=13/17 or,,,,, (52C2-(13C2*4))/52C2=13/17
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If two cards are picked out of a standard 52-card deck of cards, witho 12 Nov 2024, 00:45
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A simple and quick approach:

1 Card can be from any suit so probability of first card is 1. ( once this is done cards left in the deck will be 51)


For the 2nd card we can choose cards from 3 other suits ie from 13x3 = 39 cards.
So Probability = 39/51

P(First card ) x P(second Card) = 1x 39/51 = 13/17
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If two cards are picked out of a standard 52-card deck of cards, witho 14 Nov 2024, 01:30
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Trying to do it in another way... Bunuel please verify this
hearts, clubs, spades, diamonds
Selecting any two combination:4C2= 6
Two selected card can be arrange in 2 ways.
So probability= (13/52)*(13/51)*6*2= 13/17
Answer:B
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If two cards are picked out of a standard 52-card deck of cards, witho 14 Nov 2024, 01:53
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Vibhatu
Trying to do it in another way... Bunuel please verify this
hearts, clubs, spades, diamonds
Selecting any two combination:4C2= 6
Two selected card can be arrange in 2 ways.
So probability= (13/52)*(13/51)*6*2= 13/17
Answer:B

With this approach it should be 4C2*13C1*13C1/52C2 = 13/17.
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If two cards are picked out of a standard 52-card deck of cards, witho 14 Nov 2024, 03:07
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If I have to pick 1 card out of 52, it'd be 52C1.

Considering the remaining suites, 39C1 for the other card.

Total possibility is 52c2

So P of event : 52C1*39C1/52C2

What am I missing?
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If two cards are picked out of a standard 52-card deck of cards, witho 14 Nov 2024, 03:21
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Farhanashraf7
If I have to pick 1 card out of 52, it'd be 52C1.

Considering the remaining suites, 39C1 for the other card.

Total possibility is 52c2

So P of event : 52C1*39C1/52C2

What am I missing?

With this method, the numerator should be divided by 2. For example, using 52C1 to get the ace of spades and 39C1 to get the ace of hearts gives the pair (ace of spades, ace of hearts). However, if 52C1 gives the ace of hearts and 39C1 gives the ace of spades, you end up with the same pair (ace of spades, ace of hearts). To account for these duplicates, we should divide 52C1 * 39C1 by 2.
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If two cards are picked out of a standard 52-card deck of cards, witho 30 Jan 2025, 07:36
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If two cards are picked out of a standard 52-card deck of cards, without replacement, what is the probability that the cards picked are of two different suits ? (A standard 52-card deck of cards has four suits: hearts, clubs, spades, diamonds (13 cards in each suit).)

We can answer this question quickly by noticing that the suit of the first card does not matter. All that matters is whether the suit of the second card is different from that of the first. After all, if the suit of the second card is different from that of the first, then the cards picked are of two different suits.

So, we pick the first card. We don't have to worry about any probability in this case since the suit of the first card doesn't matter.

Then, what's the probability that the second card is of a different suit? There are 51 cards remaining, and, of those, 3 × 13 = 39 are of suits different from that of the first card.

So, the probability that the second card will be of a different suit is 39/51 = 13/17.

A. 15/17
B. 13/17
C. 11/17
D. 4/17
E. 13/102

Correct answer: B
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