Try this beautiful problem from Probability.
Problem based on Probability | AMC-8, 2016 | Problem 21
A box contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?
- \(\frac{3}{5}\)
- \(\frac{2}{5}\)
- \(\frac{1}{4}\)
Key Concepts
probability
combination
fraction
Check the Answer
But try the problem first…
Answer: \(\frac{2}{5}\)
AMC-8, 2016 problem 21
Challenges and Thrills in Pre College Mathematics
Try with Hints
First hint
There are 5 Chips, 3 red and 2 green
Can you now finish the problem ……….
Second Hint
We draw the chips boxes in such a way that we do not stop when the last chip of color is drawn. one at a time without replacement
Can you finish the problem……..
Final Step
There are 5 Chips, 3 red and 2 green
we draw the chips boxes in such a way that we do not stop when the last chip of color is drawn.
if we draw all the green chip boxes then the last box be red or if we draw all red boxes then the last box be green
but we draw randomly. there are 3 red boxes and 2 green boxes
Therefore the probability that the 3 reds are drawn=\(\frac{2}{5}\)
Other useful links
- https://www.cheenta.com/largest-and-smallest-numbers-amc-8-2006-problem-22/
- https://www.youtube.com/watch?v=_HYhADRACPs
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