The entire output of a factory is produced on three machines. The three machines account for 20%, 30%, and 50% of the output, respectively. The fraction of defective items produced is this: for the first machine, 5%; for the second machine, 3%; for the third machine, 1%.
If an item is chosen at random from the total output and is found to be defective,
what is the probability that it was produced by the third machine?

0.20833

Step-by-step explanation:

This is a case of inverse probability.

We first determine the probability that an item is defective generally.

To do this we sum the possible options of having the item produced from the first machine and is defective or having the item produced by the second machine and is defective or having the item produced by the third machine and is defective.

Mathematically,this becomes :

(0.2 * 0.05) + (0.3 * 0.03) + (0.5 * 0.01)

0.01 + 0.009 + 0.005 = 0.024.

Probability an item is defective in general = 0.024

Probability a third machine produces a defective item = 0.5 * 0.01 = 0.005.

Hence, given that the item is defective, the probability that it was produced by the third machine becomes:

= (probability of the item being produced from the third machine and is defective) / (probability of the item being defective in general)

= 0.005/0.024

= 0.20833.

do you mean add both of the equations together

because if you do the answer is 10x-5

step-by-step explanation:

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13