anonymous
  • anonymous
For this experiment,you will draw one card and replace it.Then draw one shape and replace it. Find each probability and show how much your answer. 1.What is the probability that you will draw 42 ans a star? 2.What is P(7and arrow)? 3.What is the probability that you will draw an even number and a heart? 4.What is the probability that you will not draw an odd number and not draw an arrow?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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mathteacher1729
  • mathteacher1729
Was there a picture associated with this problem?
anonymous
  • anonymous
ok hold on
anonymous
  • anonymous
16 52 42 7 16 7 3 7 42 4 The pictures are heart is 6 and star is 4 and arrow is 3

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mathteacher1729
  • mathteacher1729
I see two "sixteens" but no "Six". :(
mathteacher1729
  • mathteacher1729
Also, what are the others? Do they have shapes associated with them?
anonymous
  • anonymous
no that is all the shapes and there two 16
mathteacher1729
  • mathteacher1729
So there are 13 cards in all?
anonymous
  • anonymous
yea
mathteacher1729
  • mathteacher1729
16 52 42 7 16 7 3 7 42 4 heart is 6 star is 4 arrow is 3 So one of the threes does NOT have an arrow? Sorry to ask so many questions but this is really important in order to help you get an accurate answer.
anonymous
  • anonymous
yea
mathteacher1729
  • mathteacher1729
gotcha! Ok, I think we can start now. 1.What is the probability that you will draw 42 and a star? The key word is here is "AND" a star. So there are 13 total cards. Whatever your answer is for this question, it's going to be (SOMETHING) over (13). There are TWO -- "42" And is ONE -- "star" So that means there are THREE cards that you want (42, 42, or the star). So the answer is 3/13. Does that make sense?
anonymous
  • anonymous
it ides
mathteacher1729
  • mathteacher1729
Ok: 2.What is P(7and arrow)? This is just saying "What is the probability that you draw a seven and an arrow? It's EXACTLY like problem 1, but this time it's 7, not 42, and a heart, not an arrow. See if you can figure it out. :)
anonymous
  • anonymous
i cant

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