SOLUTION: A box has 3 red and 4 black balls. You randomly take one ball at a time without putting it back in. Stop as soon as all the black balls are drawn out of the box. How many red balls
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-> SOLUTION: A box has 3 red and 4 black balls. You randomly take one ball at a time without putting it back in. Stop as soon as all the black balls are drawn out of the box. How many red balls
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Question 1178260: A box has 3 red and 4 black balls. You randomly take one ball at a time without putting it back in. Stop as soon as all the black balls are drawn out of the box. How many red balls do you expect to be left in the box? (Hint: find the expected value of the properly defined random variable) Answer by robertb(5830) (Show Source):
You can put this solution on YOUR website! Let X = random variable equal to the number of remaining red balls.
Case 1. X = 0.
X = 0 means all 7 balls had been drawn, with black as the 7th and last ball. The number of ways of arranging the 6 previous places will be .
Case 2. X = 1.
X = 1 means 6 balls had been drawn, with black as the 6th ball (and red as the remaining ball.) The number of ways of arranging the 5 previous places will be .
Case 3. X = 2.
X = 2 means 5 balls had been drawn, with black as the 5th ball (and 2 red balls as the remaining ones.) The number of ways of arranging the 4 previous places will be
Case 4. X = 3.
X = 3 means 4 balls had been drawn, with black as the 4th ball. There is only 1 possible way of getting this scenario,
and that is when all four black balls were drawn one after the other at the very start.
The probability mass function is then as follows:
P(X=0) = 20/35
P(X=1) = 10/35
P(X=2) = 4/35
P(X=3) = 1/35
