Solving problem N4122 Bayesian probability theory

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4122. On four machines with identical and independent conditions, parts of the same name are made. On the first machine make 15%, on the second - 24%, on the third 38%, on the fourth - 23% of all parts. On the first machine, the probability of every detail to be defect-free is 0.68; for the remaining machines, these probabilities are respectively equal: 0.74; 0.89; 0.92. Find the probability that:
1) at random, the taken part will be defect-free;
2) and this part is made on the fourth machine.

Detailed solution. Decorated in Microsoft Word 2003 (Quest decided to use the formula editor)

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