anonymous
  • anonymous
A bag has 1 red marble, 4 blue marbles, and 3 green marbles. Peter draws a marble randomly from the bag, replaces it, and then draws another marble randomly. What is the probability of drawing 2 blue marbles in a row? Explain your answer.
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@Nnesha
Nnesha
  • Nnesha
oky so i guess \[\rm \frac{ blue~marbles }{ total ~number~of marbles }\]
Nnesha
  • Nnesha
ahhh i'm not good 't probability stuff ?/>

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Nnesha
  • Nnesha
@jim_thompson5910
anonymous
  • anonymous
ITS OK THANKS
jim_thompson5910
  • jim_thompson5910
@Nnesha you're on the right track
anonymous
  • anonymous
Would it be 2/8
jim_thompson5910
  • jim_thompson5910
you need to figure out how many blue marbles there are, out of how many total probability of picking blue = (# of blue)/(# total) since replacements are made, this means that the probability does not change on the second draw
jim_thompson5910
  • jim_thompson5910
probability of picking 2 blue = (probability of picking 1 blue)*(probability of picking 1 blue)
anonymous
  • anonymous
so im kinda lost...so it would be 2= 1x1?
jim_thompson5910
  • jim_thompson5910
we have 4 blue out of 8 total so you agree that the probability of picking one blue is 4/8 = 1/2 right?
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
the probability of picking 2 blue is equal to P(2 blue) = P(1 blue AND 1 blue) P(2 blue) = P(1 blue)*P(1 blue) P(2 blue) = (1/2)*(1/2) P(2 blue) = (1*1)/(2*2) P(2 blue) = 1/4
jim_thompson5910
  • jim_thompson5910
this rule only works because replacements are made and each drawing is independent
anonymous
  • anonymous
OK so what should i write down? @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
Basically those steps and the answer. But in your own words and format that's unique to you. Hopefully you see how I got that answer?
anonymous
  • anonymous
yes thank u!
jim_thompson5910
  • jim_thompson5910
you're welcome
Nnesha
  • Nnesha
thanks @jim_thompson5910 o^_^o

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