Assessment Schedule KOHIA 2014 (Statistics) BOARD GAMES: AS91585

AchievementAchievement with MeritAchievement with Excellence

Apply probability concepts in solving problems involves:selecting and using methods demonstrating knowledge of concepts and termscommunicating using appropriate representations.

Apply probability concepts, using relational thinking, in solving problems involves:selecting and carrying out a logical sequence of stepsconnecting different concepts or representations demonstrating understanding of concepts and also relating findings to a context or communicating thinking using appropriate statements.Apply probability concepts, using extended abstract thinking, in solving problems involves:devising a strategy to investigate or solve a problemidentifying relevant concepts in context developing a chain of logical reasoningmaking a statistical generalisationand also where appropriate, using contextual knowledge to reflect on the answer.

Evidence StatementOne Expected CoverageAchievement (u)Merit (r)Excellence (t)

(a)Probability of landing on the place = = = 0.45 or 45%

Probability correctly calculated.

Allow for estimation between 40% - 50%

(b)Consider the four combinations: (2,5), (3,4), (4,3) and (5,2). P(2,5) = x = P(3,4) = x = P(4,3) = x = P(5,2) = x = Using the addition rule, P(landing on Take a chance) = + + + = = 1.54%Calculation of two relevant probabilities of (2,5), (3,4), (4,3) or (5,2).

Any rounding.Correct probability calculated, with some working.

Any rounding.

(c)The most likely die roll combination is 7, because it has the most outcomes (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) as compared to any other combination. The distribution of probabilities is symmetric about the number 7. OR The mean result or expected value of one throw is 7.

Therefore, given there are 40 spaces on the board, 40 spaces 7 moves = 5.71

This approximates to about 6 turns to go around the board. Assumptions are that the probability of going to jail or being sent elsewhere by Take a Chance or Community card are small and dont have much effect on the number of turns.Since people get another throw if they roll doubles, the expected value of one turn is more than 7.The expected move is 7 whether doubles is rolled or not.

If going to jail is counted as equivalent to a move of zero, the expected number of spaces moved in a turn is

408.264 = 4.840The small chance of being moved by a card or going to jail would make the number of turns slightly higher than 4.84, but not much.The expected number of turns to go around the board is about 5. Calculates about 6 turns.Calculates about 6 turns with justification and one assumption given.Calculates about 5 turns with logical reasoning.

(d)P(rolling 3 doubles | go to jail)

= = ()3 [ + x + ()3 ] = = 0.1077 = 10.77%Calculates P(rolling three doubles), any rounding.Correct calculation with one error, any rounding.Correct answer with working, any rounding.

NN1N2A3A4M5M6E7E8

No relevant evidence.Making progress.1 of u2 of u3 of u1 of r2 of r1 t 2 of t

Two Expected CoverageAchievement (u)Merit (r)Excellence (t)

(a)CoinSpinner

YesNo

Yes18220

No415

22325

CoinSpinner

YesNo

Yes18220

No415

22325

P(SC) = 4/25 Tree, table or Venn diagram

Probability found from incorrect tree, table or Venn Diagram.

(b)P(Not used spinner Not used coin) = 1/25 = 0.04.As experimental probability is greater than zero at 0.04, the events are not mutually exclusive. The sample size is small at n = 25 but if even one person in the population has used neither spinner nor coin toss that is evidence that the two events are not mutually exclusive. Identification of relevant theory related to mutually exclusive events.Explanation of why events are not mutually exclusive with justification.

(c) P(Player two wins) = P(wins first turn) + P(wins third turn)= ( x ) + ( x x ) = P(Player one wins) = P(wins second turn) + P(Wins third turn)= ( x ) + ( x x ) = As it is equally likely for either Player one or Player two to win, Oliver could omit this section of the game and this have no effect on the long-term outcomes of the game, given it is based on luck as opposed to skill.He could choose the starter more quickly eg he could have the players flip a coin which could be a quicker, although potentially a less exciting way to start the game.

Tree, table or Venn diagram with three probabilities correctly identified

Tree, table or Venn diagram with all relevant probabilities correctly identified, or correct working allow one calculation error

Correct statement in context about equally likely probabilities, supported by working.ANDAdvice given to Oliver consistent with equally likely probabilities and sensible advice given for the game.

NN1N2A3A4M5M6E7E8

No relevant evidence.Making progress.1 of u2 of u3 of u1 of r2 of r1 of t with minor error1 of t

Three Expected CoverageAchievement (u)Merit (r)Excellence (t)

(a)P(both lose) = (1/3 x x 2/3 x x 5/6 x 7/8) = 35/576 = 0.0625Probability of one player losing correctly calculated.Correctly calculates probability. CAO

(b)P(Primula wins) = P(P wins first round) OR P(P wins the second round) OR P(P wins third round) = (x ) + (x x x )+ (x x x x 5/6 x )= 1/6 + 1/36 + 5/576 = 13/64 = 0.2031Correctly calculates probability of one winning outcome.Correctly calculates probability.

CAO

(c)P(KwinsKfirst) =(+ x x + x x x x1/6)= 53/144 = 0.3681P(KwinsKsecond) =( + + x1/6) = 223/1152 = 0.1936P(kwins)=0.3681+0.1936 = 0.5617P(k firstK wins) = P(kwinskfirst)K wins) = 0.3681/0.5617 = 0.6553Correctly calculates probability of one winning outcome.Correctly calculates probability of P(KwinsKfirst) OR P(KwinsKsecond) OR P(kwins)

CAOCorrectly calculates probability of P(k firstK wins)

CAO

(d)Model distribution of probabilities out of 100 if probabilities are as described at start of question with Katrin going first:

In the observed games Primula is winning more often than the model predicts. Even though 100 is a relatively small sample size for investigating a probability, Primula is winning 67% more often than predicted, which is a bigger difference than would be expected due to sampling variability. Katrins and Primulas estimates of their probability of winning may need to be adjusted. Calculates number or proportion of expected wins for 1, 2 and 3 arrows for either Primula or Katrin.Calculates number or proportion of expected wins for 1, 2 and 3 arrows for both Primula and Katrin, AND notes difference in probabilities for model and observations.Calculates number or proportion of expected wins for 1, 2 and 3 arrows for both Primula and Katrin, AND notes difference in probabilities for model and observations.

The size of difference between probabilities is noted and explanations for possible reasons are given wrt to sample size and sampling variability and/or change in model probabilities.

NN1N2A3A4M5M6E7E8

No relevant evidence.Making progress.1 of u2 of u3 of u1 of r2 of r1 of t 2 of t

Judgement StatementNot AchievedAchievementAchievement with MeritAchievement with Excellence

Score range0 78 1213 1819 24

AS91585 Kohia 2014