anonymous
  • anonymous
What is the probability of drawing three red cards, one at a time without replacement, from a standard deck of 52 cards?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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adajiamcneal
  • adajiamcneal
wouldnt it be 3/26 ? because theres either red or black cards which means 26 cards are red 26 are black. but they're asking for 3 different cards. im just assuming
anonymous
  • anonymous
3/52 2/17 1/8 75/676 are the options
kobeni-chan
  • kobeni-chan
adajia is right on the 3 part, but it's still talking about the chances of picking those three from the deck of 52 cards.

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kobeni-chan
  • kobeni-chan
@Catseyeglint911
adajiamcneal
  • adajiamcneal
ohhhhhh ok lol i read it as 3 cards out of the red cards lol my bad
anonymous
  • anonymous
Sorry, guys. I'm doing 3 things at once.
anonymous
  • anonymous
I know the 3 out of 26 part and the 3 out of 52 but do you know if there is something else?
kobeni-chan
  • kobeni-chan
Lol thats ok adajia. And well I would assume that the answer would be 3/52 but sometimes probability is confusing so let me look in my notes.
adajiamcneal
  • adajiamcneal
what do you mean but something else? i mean you can split it 7 different ways if you base it off of the faces meaning king queen jack spaid heart etc i agree with kobeni the answer is 3/52
kobeni-chan
  • kobeni-chan
oh waiiiitttt
anonymous
  • anonymous
So we have 26C3 and 52C3 I just didn't know if there was another combination I had to solve
adajiamcneal
  • adajiamcneal
8 ways my bad i forgot to count the ace lol
kobeni-chan
  • kobeni-chan
Half of the cards in the deck are red. So the probability of picking out one red card is 1/2. I think if you wanted to see the prob. for picking three consecutive cards like that, you would multiply the denominator by 3. So I actually think its 1/8 now.
anonymous
  • anonymous
Do you know if the "without replacement" has any effect on the answer?
adajiamcneal
  • adajiamcneal
no they just mean like if you take a card out and then place it back into the deck so no it shouldnt have any effect on it
kobeni-chan
  • kobeni-chan
Oh wait argh. I multiplied the probabilities 26/52, 25/52, and 24/52 and got 75/676. That's probably in regards to the ''w/o replacement'' because the number of red cards in the deck decreases by one each time you take one out.
kobeni-chan
  • kobeni-chan
It'll have an effect on it because if you put the card back, that's another 1/52 chance that you'll pick up that red card.
adajiamcneal
  • adajiamcneal
oh yeah duhh. lol kobeni seems to know what shes talking about so ill just let her handle it from here haha
kobeni-chan
  • kobeni-chan
aww its seriously alright lol. I had a lot of trouble with probability but I'm glad i finally figured it out c:
anonymous
  • anonymous
Wait...so wouldn't it be 26/52, 25/51, and 24/50 since cards are being taken out of the deck as well?
kobeni-chan
  • kobeni-chan
If I was the type f person to curse, this is when I would do it. That's right lol... you'll want to multiply all three of those fractions to get the *true* answer.
anonymous
  • anonymous
26/52, 25/51, and 24/50 Okay multiplying these all together I got 15600/132600 = 2/17 And don't worry about it. You're fine. :)
kobeni-chan
  • kobeni-chan
Ok o_o lol that should be right, now.
anonymous
  • anonymous
Thank you. :)

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