document.write( "Question 1051583: Can someone help me with this problem:\r
\n" );
document.write( "\n" );
document.write( "In a certain production process, 25% of the items produced are found defective. It therefore requires through inspection of all the items produced before packing them for sale.\r
\n" );
document.write( "\n" );
document.write( "a) Develop a probability density function for the number of non-defective items in a day's production of n items.\r
\n" );
document.write( "\n" );
document.write( "b) Using your answer to a), determine the probability of getting at least 50 good items if 100 items are produced.
\n" );
document.write( " \n" );
document.write( "
Algebra.Com's Answer #667054 by rothauserc(4718)![]() ![]() You can put this solution on YOUR website! We use the binary probability distribution \n" ); document.write( ": \n" ); document.write( "The probability of non-defective items is 1 - 0.25 = 0.75 \n" ); document.write( ": \n" ); document.write( "Probability(k items are non-defective) = nCk * p^k * q^(n-k), where n is number of items produced, k = number of non-defective items, p = probability of non-defective items, q = 1-p, nCk of the combination of n items taken k at a time \n" ); document.write( ": \n" ); document.write( "a) Probability(k items are non-defective) = nCk * (0.75)^k * (0.25)^(n-k) \n" ); document.write( ": \n" ); document.write( "b) Probability(at least 50 items are produced out of 100) is the summation of \n" ); document.write( "the probabilities for 50, 51, 52, ..., 100 \n" ); document.write( ": \n" ); document.write( "Probability(at least 50 items are produced out of 100) = 0.999999 \n" ); document.write( " |