Quick Probability question involving percentages Robots A & B make toys. Robot B makes 75% of toys. When made by B 5% of the toys are faulty. When made by robot A only 10% are faulty. 
Calculate the probability that when a toy is picked at random:
A) is made by robot B and is faulty 
B) is not faulty 
So far I got 
A) 0.05 x 0.75 = 0.0375 
B) 0.85 x 0.15 = 0.1275 
Is that correct or do I need to do another formula ?
 A: Your part A) result is correct. For part B), a toy is not faulty if it's made by $A$ and not faulty, or made by $B$ and not faulty, so the probability of this occurring is $0.25 \times 0.9 + 0.75 \times 0.95 = 0.9375$ instead.
Note the $0.25$ comes from $100\% - 75\% = 25\%$ of the toys being made by robot $A$, $0.9$ comes from $100\% - 10\% = 90\%$ of the toys made by robot $A$ not being faulty, the $0.75$ comes from robot $B$ making $75\%$ of the toys, and the $0.95$ comes from $100\% - 5\% = 95\%$ of the toys being made by robot $B$ not being faulty.
A: Let the following events called $A,B$ and $F$:


*

*$A$: The chosen toy is made by Robot A.

*$B$: The chosen toy is made by Robot B.

*$F$: The chosen toy is Faulty.

*$\bar{F}$: The chosen toy is not Faulty.


The answer to your questions:
A) 
is made by robot B and is faulty.
$$P[B \cap F] = P[B] P[F|B] = 0.75 \cdot 0.05 = 0.0375$$
So your answer is correct for this case.
B) 
is not faulty.
You need to use the law of total probablity, here:
$$P[\bar{F}] = P[\bar{F}|A] P[A] + P[\bar{F}|B] P[B] = 0.25 \cdot 0.9 + 0.95 \cdot 0.75 =  0.9375$$
