document.write( "Question 1082101: A coin is biased such that the head is twice as likely to occur as tail. If this coin is tossed 4 times, find
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document.write( "i- the sample space
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document.write( "ii- the probability of having at least 2 heads
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document.write( "iii- the probability of having exactly 3 heads
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document.write( "iv- the probability of having at most 4 heads. \n" );
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Algebra.Com's Answer #696177 by Boreal(15235) You can put this solution on YOUR website! HHHH \n" ); document.write( "HHHT \n" ); document.write( "HHTH \n" ); document.write( "HTHH \n" ); document.write( "THHH \n" ); document.write( "HHTT \n" ); document.write( "HTHT \n" ); document.write( "HTTH \n" ); document.write( "THHT \n" ); document.write( "THTH \n" ); document.write( "TTHH \n" ); document.write( "TTTH \n" ); document.write( "TTHT \n" ); document.write( "THTT \n" ); document.write( "HTTT \n" ); document.write( "TTTT \n" ); document.write( "Exactly two heads in 4 tosses: 6 ways of having them 4C2 \n" ); document.write( "The denominator of the product of the probabilities is always 81. The numerator numbers change in order but not in product, and they are 2*2*1*1=4. Multiply by 6 and get 24/81=8/27 \n" ); document.write( "exactly 3 heads is 4 * 2*2*2*1/87=32/81 \n" ); document.write( "exactly 4 heads is 2*2*2*2/81=16/81 \n" ); document.write( "(and 1 head and 0 heads for completeness: \n" ); document.write( "1:2*4/81=8/81 \n" ); document.write( "0:1/81 \n" ); document.write( "Sum is 81/81 \n" ); document.write( " |