Write down the total number of fish in the pond.

Leanne catches a fish.

Find the probability that she

(i) catches an undersized bream;

(ii) catches either a flathead or an undersized fish or both;

(iii) does not catch an undersized whiting;

(iv) catches a whiting given that the fish was normal.

Leanne notices that on windy days, the probability she catches a fish is 0.1 while on non-windy days the probability she catches a fish is 0.65. The probability that it will be windy on a particular day is 0.3.

Copy and complete the probability tree diagram below.

Leanne notices that on windy days, the probability she catches a fish is 0.1 while on non-windy days the probability she catches a fish is 0.65. The probability that it will be windy on a particular day is 0.3.

Calculate the probability that it is windy and Leanne catches a fish on a particular day.

Leanne notices that on windy days, the probability she catches a fish is 0.1 while on non-windy days the probability she catches a fish is 0.65. The probability that it will be windy on a particular day is 0.3.

Calculate the probability that Leanne catches a fish on a particular day.

Use your answer to part (e) to calculate the probability that Leanne catches a fish on two consecutive days.

Leanne notices that on windy days, the probability she catches a fish is 0.1 while on non-windy days the probability she catches a fish is 0.65. The probability that it will be windy on a particular day is 0.3.

Given that Leanne catches a fish on a particular day, calculate the probability that the day was windy.

90     (A1)

[1 mark]

(i) \(\frac{3}{{90}}(0.0\bar 3,{\text{ }}0.0333,{\text{ }}0.0333...,{\text{ }}3.\bar 3\% ,{\text{ }}3.33\% )\)     (A1)(ft)

Note: For the denominator follow through from their answer in part (a).

(ii) \(\frac{{53}}{{90}}(0.5\bar 8,{\text{ }}0.588...,{\text{ }}0.589,{\text{ }}58.\bar 8\% ,{\text{ }}58.9\% )\)     (A1)(A1)(ft)(G2)

Notes: Award (A1) for the numerator. (A1)(ft) for denominator. For the denominator follow through from their answer in part (a).

(iii) \(\frac{{72}}{{90}}{\text{(0.8, 80}}\%)\)     (A1)(ft)(A1)(ft)(G2)

Notes: Award (A1)(ft) for the numerator, (their part (a) –18) (A1)(ft) for denominator. For the denominator follow through from their answer in part (a).

(iv) \(\frac{{24}}{{48}}(0.5,{\text{ 50}}\% )\)     (A1)(A1)(G2)

Note: Award (A1) for numerator, (A1) for denominator.

[7 marks]

     (A1)(A1)(A1)

Notes: Award (A1) for each correct entry. Tree diagram must be seen for marks to be awarded.

[3 marks]

\(0.3 \times 0.1 = 0.03\left( {\frac{3}{{100}}} \right)\)     (M1)(A1)(G2)

Note: Award (M1) for correct product seen.

[2 marks]

\(0.3 \times 0.1+ 0.7\times0.65\)     (M1)(M1)

Notes: Award (M1) for \(0.7\times0.65\) (or 0.455) seen, (M1) for adding their 0.03. Follow through from their answers to parts (c) and (d).

\( = 0.485\left( {\frac{{485}}{{1000}},\frac{{97}}{{200}}} \right)\)     (A1)(ft)(G2)

Note: Follow through from their tree diagram and their answer to part (d).

[3 marks]

\(0.485 \times 0.485\)     (M1)

\(0.235\left( {\frac{{9409}}{{40000}}{\text{, }}0.235225} \right)\)     (A1)(ft)(G2)

Note: Follow through from their answer to part (e).

[2 marks]

\(\frac{{0.03}}{{0.485}}\)     (M1)(A1)(ft)

Notes: Award (M1) for substituted conditional probability formula, (A1)(ft) for their (d) as numerator and their (e) as denominator.

\(0.0619\left( {\frac{{6}}{{97}}}\text{, 0.0618556...} \right) \)     (A1)(ft)(G2)

Note: Follow through from their parts (d) and (e).

[3 marks]

(a) Most candidates found this correctly although a few wrote 180 instead of 90.

(b) This was also answered well. The main errors were putting 65/90 in part (ii) and 24/90 in part (iv).

(c) The tree diagram was completed correctly in most scripts. It appears that some candidates may have answered this on their question paper and this was not sent to the scanning centre with the answer papers.

(d) Many answered this correctly. Some added instead of multiplying.

(e) Surprisingly well answered. Again some added and multiplied in the wrong place.

(f) Most candidates added here and then divided by 2 rather than multiplying.

(g) This was badly done with very few correct answers seen.