One of the dogs is chosen at random.

(i) Find P (the dog is grey and has the yellow bowl).

(ii) Find P (the dog does not get the yellow bowl).

Neil often takes the dogs to the park after they have eaten. He has noticed that the grey dog plays with a stick for a quarter of the time and both brown dogs play with sticks for half of the time. This information is shown on the tree diagram below.

(i) Copy the tree diagram and add the four missing probability values on the branches that refer to playing with a stick.

During a trip to the park, one of the dogs is chosen at random.

(ii) Find P (the dog is grey or is playing with a stick, but not both).

(iii) Find P (the dog is grey given that the dog is playing with a stick).

(iv) Find P (the dog is grey and was fed from the yellow bowl and is not playing with a stick).

Complete the diagram, using the information given in the question.

Find (i) \(n(M \cap W)\)

(ii) \(n(M′ \cup S)\)

Two mice are chosen without replacement.

Find P (both mice are short-tailed).

(i) P (a dog is grey and has the yellow bowl)

\( = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}( = 0.111)\)     (M1)(A1)(G2)

The (M1) is for multiplying two values along any branch of the tree.

(ii) P (dog does not get yellow bowl) \( = \frac{2}{3}\) ( = 0.667 (3sf) or 0.6)     (A1)

[3 marks]

(i) The tree diagram should show the values \(\frac{1}{2},\frac{1}{2}\) for the brown branch and \(\frac{1}{4},\frac{3}{4}\) in the correct positions for the grey branch.     (A1)(A1)(ft)

Follow through if the branches are interchanged.

(ii) P (the dog is grey or is playing with a stick, but not both)

\( = \frac{1}{3} \times \frac{3}{4} + \frac{2}{3} \times \frac{1}{2}\)     (M1)

\( = \frac{7}{{12}}\)     ( = 0.583)     (A1)(ft)(G1)

The (M1) is for showing two correct products (whether added or not). Follow through from b(i). Award (M1) for \( \frac{1}{3} + \frac{1}{4}\) (must be a sum).

(iii) P (dog is grey given that it is playing with stick)

\(\frac{{P(G \cap S)}}{{P(S)}} = \frac{{\frac{1}{3} \times \frac{1}{4}}}{{\left( {\frac{2}{3} \times \frac{1}{2}} \right) + \left( {\frac{1}{3} \times \frac{1}{4}} \right)}}\) or \(\frac{1}{{12}}/\frac{5}{{12}}\)     (M1)(A1)(ft)

(M1) for substituted conditional probability formula, (A1) for correct substitutions.

\( = \frac{1}{5}\)    ( = 0.2)     (A1)(ft)(G2)    

(iv) P (grey and fed from yellow bowl and not playing with stick) \( = \frac{1}{3} \times \frac{1}{3} \times \frac{3}{4} = \frac{1}{{12}}\)   ( = \(\frac{3}{{36}}\) = 0.0833 3sf).     (M1)(A1)(ft)(G1)

(M1) is for product of 3 reasonable probability values.

[9 marks]

     (A1)(A1)(A1)(ft)(A1)(ft)

Award (A1) for 2 (must be in a box), (A1) for 7, (A1)(ft) for 6 and 4, (A1)(ft) for 9 and 13. Observe the assignment of (ft) marks strictly here. Example A common error is likely to be 11 instead of 6 (A0). In this case follow through to 4 and 18 (A1)(ft) for the final pair. Here the 4 follows from the total of 27 for n(M).

[4 marks]

(i) \(n(M \cap W) = 14\)     (A1)(ft)

(ii) \(n(M' \cup S) = 22 + 11\) OR \(15 + 18\)     (A1)(ft)

= 33     (A1)(ft)

Award (A2) if answer 33 is seen. Award (A1) for any of 22, 11, 15 or 18 seen but 33 absent.

[3 marks]

P (both mice short-tailed) \( = \frac{{18}}{{49}} \times \frac{{17}}{{48}} = \frac{{306}}{{352}}\) (= 0.130).     (M1)(A1)(ft)(G1)

(Allow alternatives such as 153/1176 or 51/392.) Award (M1) for any of \(\frac{{18}}{{49}}\) and \(\frac{{17}}{{48}}\) or \(\frac{{18}}{{49}} \times \frac{{17}}{{49}}\) or \(\frac{{18}}{{49}} + \frac{{17}}{{48}}\) seen.

[2 marks]

(i) (a),(b) Elementary probability calculations were performed well and compound ones often poorly. Filling in of the tree diagram in b(i) was quite well done. Conditional probability in particular was poorly implemented.

(i) (a),(b) Elementary probability calculations were performed well and compound ones often poorly. Filling in of the tree diagram in b(i) was quite well done. Conditional probability in particular was poorly implemented.

(ii) Most candidates had some idea how to fill in the numbers on the diagram. Full marks were common here and most candidates got some of the marks.

 Part b(i) was handled better than b(ii), with the complement causing problems. Extensive follow-through was used here from (a).

Part (c) was rarely completed, perhaps due to time constraints, but also due to lack of understanding.