anonymous
  • anonymous
Two balls are drawn in succession out of a box containing 5 red and 5 white balls. Find the probability that at least 1 ball was red, given that the first ball was.. A) Replaced before the second draw B) NOT replaced before the second draw a) Find the probability that at least 1 ball was red, given that the first ball was replaced before the second draw.
Probability
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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goformit100
  • goformit100
I know the solution...
anonymous
  • anonymous
Can you please help me?
kropot72
  • kropot72
Section A) Consider the following 3 situations: (1) A: First ball red B: Second ball red The events A and B are independent therefore P(two red) = (5/10) * (5/10) ...........................(1) (2) A: First ball red B: Second ball black The events A and B are independent therefore P(one red) = (5/10) * (5/10) ...........................(2) (3) A: First ball black B: Second ball red The events A and B are independent therefore P(one red) = (5/10) * (5/10) ...........................(3) The probability that at least 1 ball was red, given that the first ball was replaced before the second draw is the sum of results (1), (2) and (3). Can you calculate?

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More answers

anonymous
  • anonymous
Sorry i can not calculate it,
anonymous
  • anonymous
my calculator is currently not working
anonymous
  • anonymous
so this is what it would look like then (5/10)(5/10)+(5/10)(5/10)+(5/10)(5/10)=
kropot72
  • kropot72
You do not need a calculator to work this out: \[P(at\ least\ one\ red)=(\frac{5}{10}\times \frac{5}{10})+(\frac{5}{10}\times \frac{5}{10})+(\frac{5}{10}\times \frac{5}{10})=?\]
kropot72
  • kropot72
Hint: \[\frac{5}{10}\times \frac{5}{10}=\frac{1}{2}\times \frac{1}{2}\]
anonymous
  • anonymous
so the answer would be 0.75
kropot72
  • kropot72
Good work! Your answer is correct.
anonymous
  • anonymous
oops no sorry my mistake
kropot72
  • kropot72
0.75 or 3/4 is correct.
anonymous
  • anonymous
How would i find the probability that at least 1 ball was red, given that the first ball s not replaced before the second draw?
anonymous
  • anonymous
take the chance after by subtracting out already drawn balls
anonymous
  • anonymous
hmm?
kropot72
  • kropot72
Section B) Probability first ball is red = 5/10 Probability second ball is red = 4/9 P(two red) = (5/10 * 4/9) ................(1) Probability first ball is red = 5/10 Probability second ball is black = 5/9 P(one red) = (5/10 * 5/9) ...............(2) Probability first ball is black = 5/10 Probability second ball is red = 5/9 P(one red) = (5/10 * 5/9) ...............(3) \[P(at\ least\ one\ red)=(\frac{1}{2}\times \frac{4}{9})+(\frac{1}{2}\times \frac{5}{9})+(\frac{1}{2}\times \frac{5}{9})=?\]
anonymous
  • anonymous
thank you and the answer i have recieved was 0.77777777777
anonymous
  • anonymous
is that also what you got?
kropot72
  • kropot72
Your answer to B) is correct. Good work :)
anonymous
  • anonymous
thank you for your help! :)
anonymous
  • anonymous
but should i put my answer as 0.78?
anonymous
  • anonymous
it says to simply my answer into a integer or fraction
anonymous
  • anonymous
simplify**
anonymous
  • anonymous
oops nvm its 7/9 hahaha thanks!
kropot72
  • kropot72
To put the answer as a fraction just add the following fractions and simplify: \[\frac{4}{18}+\frac{5}{18}+\frac{5}{18}=?\]
kropot72
  • kropot72
Yes, 7/9 is correct!
anonymous
  • anonymous
:)
kropot72
  • kropot72
You're welcome :)

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