
Test solutions with the detailed point scale. Add up your points and spot what to review.
1 Counting / 3 pts
(a) Order matters: \(P(9,3)=9\times8\times7=504\). (1 pt)
(b) Order does not matter: \(C(9,3)=\dfrac{504}{6}=84\). (1 pt)
(c) 7 letters with L twice and O twice: \(\dfrac{7!}{2!\,2!}=\dfrac{5040}{4}=1260\). (1 pt)
2 Marbles / 4 pts
(a) \(\dfrac5{10}\times\dfrac49=\dfrac{20}{90}=\dfrac29\). (1.5 pts)
(b) Red: \(\dfrac{20}{90}\). Blue: \(\dfrac3{10}\times\dfrac29=\dfrac6{90}\). Yellow: \(\dfrac2{10}\times\dfrac19=\dfrac2{90}\). Sum: \(\dfrac{28}{90}=\dfrac{14}{45}\). (1.5 pts)
(c) The draws are independent: \(\dfrac5{10}\times\dfrac5{10}=\dfrac14\). (1 pt)
3 Conditional probability / 3 pts
(a) \(\dfrac{36}{80}=0.45\). (1 pt)
(b) There are \(36+42=78\) purchases: \(\dfrac{36}{78}=\dfrac6{13}\approx0.462\). (1 pt)
(c) \(P(\text{purchase})=\dfrac{78}{150}=0.52\neq0.45\), so the events are not independent. (1 pt)
4 Binomial model / 3 pts
(a) \(C(6,5)(0.9)^5(0.1)=6\times0.59049\times0.1\approx0.3543\). (1 pt)
(b) Add \(P(X=6)=0.9^6\approx0.5314\): total \(\approx0.8857\). (1 pt)
(c) Mean \(=6\times0.9=5.4\); \(\sigma=\sqrt{6\times0.9\times0.1}=\sqrt{0.54}\approx0.735\). (1 pt)
5 Watermelons / 4 pts
(a) \(z=\dfrac{15-12}{1.5}=2\) (1 pt) and \(z=\dfrac{10.5-12}{1.5}=-1\). (1 pt)
(b) 9 and 15 are \(\mu\pm2\sigma\): about 95%. (1 pt)
(c) 13.5 is \(\mu+\sigma\): \(\dfrac{100-68}{2}=16\%\). (1 pt)
6 Poll and bias / 3 pts
(a) \(ME=1.96\sqrt{\dfrac{0.54\times0.46}{900}}\approx1.96\times0.0166\approx0.033\), about 3.3%. (1 pt)
(b) The interval is about 50.7% to 57.3%. It lies entirely above 50%, so a majority is supported at the 95% level. (1 pt)
(c) It is a convenience sample: gym users are not representative of the whole population, so the result is biased no matter how many people answer. (1 pt)
Test yourself: quick challenge for Grade 11
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