anonymous
  • anonymous
This question is related to probability, please help ASAP!!! Bag A contains 3 red balls and 3 blue balls. Bag B contains 1 red ball and 3 blue balls. Q: a ball is taken at random from bag A and placed in bag B. Then a ball is chosen from bag B. What is probability that the ball taken from bag B is red... Explain the logic... i know the answer but i dont know the logic behind it please help
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
What's the probability that the ball you remove is red if you added a red ball from bag A?
anonymous
  • anonymous
2/5
anonymous
  • anonymous
am i right? @Jemurray3

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anonymous
  • anonymous
Yes, that's right. What's the probability that the ball you remove is red if you added a blue ball from bag A?
anonymous
  • anonymous
1/5
anonymous
  • anonymous
Lastly, what's the probability that you picked a red marble from bag A? Or a blue one?
anonymous
  • anonymous
1/2
anonymous
  • anonymous
or 3/6
anonymous
  • anonymous
Okay. So the two ways you can end up with a red marble at the end are: (1) Move a red one from bag A to bag B, and then pick a red one from bag B (2) Move a blue one from bag A to bag B, and then pick a red one from bag B
anonymous
  • anonymous
yes
anonymous
  • anonymous
The probability of the first one is (1/2) * (2/5) and the probability for the second one is (1/2) * (1/5). Since both sequences give us a red one at the end, you add the probabilities to get the final answer.
anonymous
  • anonymous
yea i know that but the part i didnt understand is why are we multiplying 1/2 with each possibility
anonymous
  • anonymous
Because that's the probability of choosing a red or blue marble from bag A to begin with.
anonymous
  • anonymous
ok its is but why multiplying
anonymous
  • anonymous
@cwrw238 read and just answer my last comment plz
anonymous
  • anonymous
@jim_thompson5910 read and just answer my last comment plz
anonymous
  • anonymous
If you want the probability that you have event A and also event B, you multiply the respective probabilities.
anonymous
  • anonymous
ohh thanx it was very helpful @Jemurray3

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