Probability question on drawing cards

micky123

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From a standard pack of 52 cards, I draw 3 cards at random without replacement. What is the probability that all cards drawn are both hearts and picture cards?

is the answer: (3/52)*(2/51)*(1/50)?
 

Subhotosh Khan

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From a standard pack of 52 cards, I draw 3 cards at random without replacement. What is the probability that all cards drawn are both hearts and picture cards?

is the answer: (3/52)*(2/51)*(1/50)?
There are 4 cards (ACE included) with pictures AND heart.

How many ways can you draw 3 cards from that group ?

How many ways can you draw 3 cards from 52 cards ?
 

micky123

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Why would ace be included? I thought there are only 3 cards that are both heart and picture? (jack, queen and king)

Please kindly explain if I have gotten it wrong! Thank you.
 

Subhotosh Khan

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Why would ace be included? I thought there are only 3 cards that are both heart and picture? (jack, queen and king)

Please kindly explain if I have gotten it wrong! Thank you.
If ACE is not included then - how many ways can you choose 3 cards from that group (Hearts and pictures) ? ------ 1 way

How many ways can you choose 3 cards (no restriction) from the deck? _____ \(\displaystyle C^{52}_{3}\)

Now calculate probability.

reference: https://math.stackexchange.com/ques...-getting-3-cards-in-the-same-suit-from-a-deck
 

micky123

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If ace is not included, then the probability is 1/22100. However, i am confused as to why there are 4 cards with pictures and heart?

Am I understanding the event incorrectly? I understood the event that “all three cards are both heart and picture” to be drawing “a jack, a queen and a king of hearts”, which is 1 out of 22100 ways.
 

pka

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If ace is not included, then the probability is 1/22100. However, i am confused as to why there are 4 cards with pictures and heart?

Am I understanding the event incorrectly? I understood the event that “all three cards are both heart and picture” to be drawing “a jack, a queen and a king of hearts”, which is 1 out of 22100 ways.
I would agree with your first answer [imath]\left(\dfrac{3}{52}\right)\cdot\left(\dfrac{2}{51}\right)\cdot\left(\dfrac{1}{50}\right)[/imath].
BUT if I were reading this question as an editor I would object to its wording as being unclear. It is extremely unusual to use picture card what standard English uses face cards.
 

Otis

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confused as to why there are 4 cards with pictures and heart?
Because a lot of card manufacturers draw a picture on each ace. However, the design on the aces do not show a person.

The name 'picture card' refers only to cards containing a picture of a person.

😎
 

micky123

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I see. So this means that it depends on the way “picture card” is defined in the question?

if ace is not considered a picture card, the ans would be (3/52)*(2/51)*(1/50).

if ace is considered a picture card, the and would be (4/51)*(3/51)*(2/50).

am i right?
 
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