Pr (10 and a heart) - Am I overthinking this?

#1
Hi guys,

This may be way simplistic for this forum, so I apologise in advance.

Drawing two cards from a standard 52 card deck. What is the possibility of drawing a 10 and a heart?


So, the 10 of hearts changes things because they are dependent events.

So it's not simply 4/52*13/51

The possible draws involve draw 1 = 10 hearts and event 2 is 15/51 10's or hearts

draw 1 heart (not 10) 12/52 and draw 2: 4/51 10's

Draw 1 a 10 = 3/52 (already did 10 hearts) and 13/51 (hearts)

Simply add these probabilities? Is my logic flawed?

Seriously, ignore all my babbling, can anyone explain this simply for me.
 
#2
The total number of way to draw 2 cards from a deck of 52 is \({52}\choose{2}\) = 1326

There are 51 winning hands that have a 10 of hearts(a ten of hearts and any other card.)

There are 3*12 winning hands that do not have the ten of hearts. The other 3 tens * the other 12 hearts.

\(\frac{51 + 36}{1326} = \frac{87}{1326} \approx 0.0656 \)
 
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#3
Thank you for your response, but I am still unsure.

Wouldn't the number of potential hands be 1/52 x 1/51. We can halve that if we do not care about order?

Therefore 1326?

Or must this be calculcated with combinations/permutations stuff.
 
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