document.write( "Question 1117538: Hello, I have a question! I tried the first one on my own and I think it's correct.Any help is greatly appreciated! Thank you so much! It helps me out a lot!\r
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document.write( "It's a two part question:\r
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document.write( "I flip a coin 3 times and get all heads. \r
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document.write( "1. Does that mean the next has got to be tails? Why or why not. I said no, because the next time you flip, it could be heads again. \r
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document.write( "2. In the long run, what proportion of flips will up as heads? This is where I am stuck. Again any help is greatly appreciated and welcomed! Thank you so much! \n" );
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Algebra.Com's Answer #732642 by solver91311(24713)![]() ![]() You can put this solution on YOUR website! \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Your answer to part 1 is absolutely correct. If you flip a (fair) coin 99 times and get a head each time, the probability of a tail on the 100 flip is still one-half, exactly what the probability was on each of the preceding flips.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "With a fair coin, as in one that doesn't have a drop of solder on it to make one side a little heavier, the probability of a specific outcome, let's say a head, on any given trial is one-half. You have one outcome that you call a success out of two possible outcomes. Since either outcome is equally likely (c.f. \"fair coin\"), the long run expectation is getting a head half of the time and a tail the other half.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "However, do not confuse the concept of \"expected outcome\" with \"predicting a future outcome\". Every trial is completely independent of any previous trial or trials.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "John \n" ); document.write( " \n" ); document.write( "My calculator said it, I believe it, that settles it \n" ); document.write( " ![]() \n" ); document.write( " |