My simple easy homework question. Just needed some double check :D
Deal 4 cards from a deck of 52 cards. What is the probability that we get one card from each suit?
My answer
First Draw: We can get any card, and the card's suit will be done. $Chance:1$
Second Draw: Now we need to get 1 of the 3 remaining suits. There are 51 cards left. $Chance:\frac{13+13+13}{51}$
Third Draw: Now we need to get 1 of the 2 remaining suits. There are 50 cards left. $Chance:\frac{13+13}{50}$
Fourth Draw: Now we need to get the last remaining suit. There are 49 cards left. $Chance:\frac{13}{49}$
$P($One card from each suit$)=1*\frac{13+13+13}{51}*\frac{13+13}{50}*\frac{13}{49}=0.1055$
My tutor is known for giving not-so straightforward questions, so I'm wondering if I need to consider another way, or I could be wrong. Any alternatives welcome too!