document.write( "Question 1177712: 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) \n" ); document.write( "
Algebra.Com's Answer #850400 by CPhill(1987) You can put this solution on YOUR website! Let's solve this problem by considering the probability of each red ball remaining in the box.\r \n" ); document.write( "\n" ); document.write( "**Understanding the Problem**\r \n" ); document.write( "\n" ); document.write( "We have 3 red balls and 4 black balls. We draw balls without replacement until all 4 black balls are drawn. We want to find the expected number of red balls remaining.\r \n" ); document.write( "\n" ); document.write( "**Reasoning**\r \n" ); document.write( "\n" ); document.write( "A red ball will remain in the box if it is drawn *after* all the black balls have been drawn. This means that a red ball remains if the last black ball is drawn *before* that red ball is drawn.\r \n" ); document.write( "\n" ); document.write( "Consider the positions of the balls in the sequence of draws. We have a total of 7 balls (3 red, 4 black).\r \n" ); document.write( "\n" ); document.write( "Let's look at a specific red ball. It will remain in the box if the last black ball is drawn before it.\r \n" ); document.write( "\n" ); document.write( "Consider any specific red ball and the last black ball. There are two possibilities:\r \n" ); document.write( "\n" ); document.write( "1. The last black ball is drawn before the red ball. \n" ); document.write( "2. The red ball is drawn before the last black ball.\r \n" ); document.write( "\n" ); document.write( "Since the draws are random, these two possibilities are equally likely. Therefore, the probability that a specific red ball remains in the box is 1/2.\r \n" ); document.write( "\n" ); document.write( "**Expected Value**\r \n" ); document.write( "\n" ); document.write( "Let X be the number of red balls remaining in the box. We have 3 red balls. \n" ); document.write( "* E(X) = (probability that red ball 1 remains) + (probability that red ball 2 remains) + (probability that red ball 3 remains) \n" ); document.write( "* E(X) = 1/2 + 1/2 + 1/2 = 3/2 = 1.5\r \n" ); document.write( "\n" ); document.write( "**Therefore, the expected number of red balls remaining in the box is 1.5.** \n" ); document.write( " \n" ); document.write( " |