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If I choose four cards from a standard $52$-card deck, with replacement, what is the probability that I will end up with one card from each suit?

 Oct 15, 2020
 #1
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+3

I think.

 

\(\text{P(each suit different)}=1* \frac{39*26*13}{52*52*52}\\ \text{P(each suit different)}= \frac{3}{4*2*4}\\ \text{P(each suit different)}= \frac{3}{32}\\ \)

 Oct 15, 2020
 #2
avatar+13 
0

Correctlaugh

MathBoyFarcastik  Oct 15, 2020

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