A tailor buys buttons from one of two suppliers. Supplier A's box has 3 defective and 9 good; Supplier B's has 2 defective and 8 good. Two buttons are picke...

This question tests reading the complement of a known branch probability, probability without replacement using a tree diagram, the "at least one" complement technique, and comparing probabilities between two independent tree diagrams.

(a) The two branches from the "Good" node (good then good, good then defective) must have probabilities summing to 1, since after removing one good button, the second button from Supplier A's box is either good or defective:

\[ x = 1 - \frac{3}{11} = \frac{8}{11} \] [1 mark]

(b) Multiply along the "good then good" path of the tree:

\[ P(\text{both good, A}) = \frac{3}{4} \times \frac{8}{11} = \frac{24}{44} = \frac{6}{11} \] [2 marks]

(c) "At least one defective" is the complement of "both good" found in part (b):

\[ P(\text{at least one defective, A}) = 1 - \frac{6}{11} = \frac{5}{11} \] [1 mark]

(d) Compare the two suppliers' probabilities of giving two good buttons as decimals: \( \frac{6}{11} \approx 0.545 \) for Supplier A and \( \frac{28}{45} \approx 0.622 \) for Supplier B. Since \( 0.622 \) is greater than \( 0.545 \), Supplier B is more likely to give the tailor two good buttons. [2 marks]

Comparing two fractions with different denominators is most reliably done by converting both to decimals, as in part (d), rather than trying to compare \( \frac{6}{11} \) and \( \frac{28}{45} \) directly.