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1. Jane plays a game which involves two coins being tossed. The amounts to be won for the different possible outcomes are shown below: Win €6 for two heads Win €1 for one head and one tail Win €2 for two tails It costs €4 to play one game. 1.

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  1. 1. Jane plays a game which involves two coins being tossed. The amounts to be won for the different possible outcomes are shown below: Win €6 for two heads Win €1 for one head and one tail Win €2 for two tails It costs €4 to play one game.

  2. 1. Jane plays a game which involves two coins being tossed. The amounts to be won for the different possible outcomes are shown below: Win €6 for two heads Win €1 for one head and one tail Win €2 for two tails It costs €4 to play one game. (i) Will Jane expect a gain or a loss, and how much will it be? Justify your answer with suitable calculations. Costs €4 to play €2·50 – €4 = €1·50 loss Jane will expect a loss of €1·50.

  3. 1. Jane plays a game which involves two coins being tossed. The amounts to be won for the different possible outcomes are shown below: Win €6 for two heads Win €1 for one head and one tail Win €2 for two tails It costs €4 to play one game. (ii) Is this a fair game? Justify your answer. It is not a fair game as the player will expect a loss. It is unfair for the player.

  4. 2. Conor says that the probability of getting two tails in three throws is higher than the probability of getting three heads in a row. (i) Do you agree with Conor? Justify your answer by showing the calculations for each situation. P(Two Tails in three throws) = P(TTH) or P(THT) or P(HTT) = = =

  5. 2. Conor says that the probability of getting two tails in three throws is higher than the probability of getting three heads in a row. (i) Do you agree with Conor? Justify your answer by showing the calculations for each situation. P(HHH) = = Conor is right and it is more likely to get 2 tails in 3 throws than 3 heads.

  6. 2. Georgia says that if you get three heads in a row you are more likely to get a tail on the next throw because a tail is due. (ii) Do you agree with Georgia? Explain your answer. No, I don’t agree with Georgia. Flipping a coinis an independentevent. The previousthrowdoes not effectthenext. The probability of getting a tail will always be .

  7. 3. The table shows the probability of choosing each of the different coloured sweets from packets containing red, green, yellow, brown and blue sweets. (i) What is the probability of choosing a yellow sweet? Probabilities always add up to 1. 1 – (0·3 + 0·15 + 0·26 + 0·18) = 1 – 0·89 = 0·11 = P(yellow)

  8. 3. The table shows the probability of choosing each of the different coloured sweets from packets containing red, green, yellow, brown and blue sweets. (ii) What is the probability of not choosing a red sweet? P(Not red) = 1 – P(red) = 1 – 0·3 = 0·7

  9. 3. The table shows the probability of choosing each of the different coloured sweets from packets containing red, green, yellow, brown and blue sweets. (ii) If three sweets are chosen at the same time, what is the probability of choosing a green, a blue and a brown sweet? You may have solved this question by multiplying the three probabilities, however this is incorrect. Once one sweet is removed, the total number of sweets remaining changes and so the probability of selecting a certain colour also changes. These events are dependent. Therefore, without knowing the number of each coloured sweet present at the start, we cannot calculate the probabilities after each sweet is selected.

  10. 4. Weather records for a town suggest the following: If it rains (R) on a given day, the probability it will rain the next day is . If it is sunny (S) on a given day, the probability of the next day being sunny is . In a particular week Tuesday is sunny. The tree diagram shows the possible outcomes for the next three days. (i) Show that the probability of Thursday being sunny is . P(Thursday sunny) = P(SS) or P(RS) = = =

  11. 4. Weather records for a town suggest the following: If it rains (R) on a given day, the probability it will rain the next day is . If it is sunny (S) on a given day, the probability of the next day being sunny is . In a particular week Tuesday is sunny. The tree diagram shows the possible outcomes for the next three days. (ii) What is the probability of rain on both Thursday and Friday? P(RR) = cannot ignore Wednesday ∴ P(RRR) or P(SRR) = = =

  12. 4. Weather records for a town suggest the following: If it rains (R) on a given day, the probability it will rain the next day is . If it is sunny (S) on a given day, the probability of the next day being sunny is . (iii) What is the probability of at least one of Thursday and Friday being sunny? P(At least one of Thurs/Fri Sunny) = P(RRS) or P(RSR) or P(RSS) or P(SSS) or P(SRS) or P(SSR) = = =

  13. 5. Judy is playing a game in which she rolls a fair dice three times and tries to get ‘6’ as many times as she can. Using the options ‘6’ and ‘not a 6’, draw a treediagram to representthisgame. Fill in the probabilities on the branches of the treediagram. (i)

  14. 5. Judy is playing a game in which she rolls a fair dice three times and tries to get ‘6’ as many times as she can. Calculate the probabilitythat Judy gets a ‘6’ on all three tries. (ii) P(666) = =

  15. 5. Judy is playing a game in which she rolls a fair dice three times and tries to get ‘6’ as many times as she can. Calculate the probabilitythat Judy gets a ‘6’ on at least two of herthree tries. (iii) P(At least 2 6s) = P(6, 6, Not 6) or P(6, Not 6, 6) or P(666) or P(Not 6, 6, 6) = = = =

  16. 6. A bag contains 10 counters with the letters of the word STATISTICS written on them. A counter is chosen at random and not replaced before choosing another one. (i) Draw a tree diagram to represent this situation. S, T, C = consonants = 7 letters Tree Diagram A, I = Vowels = 3 letters

  17. 6. A bag contains 10 counters with the letters of the word STATISTICS written on them. A counter is chosen at random and not replaced before choosing another one. (ii) What is the probability of getting two consonants? P(2 consonants) = P(SS) or P(ST) or P(SC) or P(TS) or P(TT) or P(TC) or P(CC) or P(CT) or P(CS) = = = =

  18. 6. A bag contains 10 counters with the letters of the word STATISTICS written on them. A counter is chosen at random and not replaced before choosing another one. (ii) What is the probability of getting two consonants? Or P(consonant, consonant) = = =

  19. 6. A bag contains 10 counters with the letters of the word STATISTICS written on them. A counter is chosen at random and not replaced before choosing another one. (iii) What is the probability of getting a vowel at least once? P(1 vowel at least) = P(vowel, vowel) or P(vowel, consonant) or P(consonant, vowel) = = = =

  20. 6. A bag contains 10 counters with the letters of the word STATISTICS written on them. A counter is chosen at random and not replaced before choosing another one. (iv) What is the probability of getting exactly one vowel? P(1 vowel) = P(vowel, consonant) or P(consonant, vowel) = = =

  21. 6. A bag contains 10 counters with the letters of the word STATISTICS written on them. A counter is chosen at random and not replaced before choosing another one. (v) What is the probability of not getting exactly one vowel? P(Not 1 vowel) = 1 – P(1 vowel) = =

  22. 7. P(success) = P(Heart) In a certain game, you pick a card from a standard deck of 52 cards. If the card is a heart, you win. If the card is not a heart, you replace the card to the deck, reshuffle, and draw again. What is the probability that you will pick the first heart on the third draw? = P(Failure) = 1 – P(Success) P(Failure) = = P(not heart and not heart and heart) = P(FFS) P(FFS) = =

  23. 8. A biased die is used in a game. The probabilities for each outcome are given below. (i) The die isthrown once. Whatis the probability of getting: (a) an even number? P(Even) = P(2) or P(4) or P(6) = 0·12 + 0·14 + 0·09 = 0·35

  24. 8. A biased die is used in a game. The probabilities for each outcome are given below. (i) The die isthrown once. Whatis the probability of getting: (b) a prime number or an odd number? P(Prime or Odd) = P(prime) + P(odd) – P(both) [Not Mutually exclusive] = P(2, 3, 5) + P(1, 3, 5) – P(3, 5) = (0·12 + 0·25 + 0·18) + (0·22 + 0·25 + 0·18) – (0·25 + 0·18) = 0·55 + 0·65 – 0·43 = 0·77

  25. 8. A biased die is used in a game. The probabilities for each outcome are given below. (i) The die isthrown once. Whatis the probability of getting: (c) a 4 or less? P(A 4 or less) = P(1, 2, 3, or 4) = 0·22 + 0·12 + 0·25 + 0·14 = 0·73

  26. 8. A biased die is used in a game. The probabilities for each outcome are given below. (i) The die isthrown once. Whatis the probability of getting: (d) a multiple of 3? P(A multiple of 3) = P(3 or 6) = 0·25 + 0·09 = 0·34

  27. 8. A biased die is used in a game. The probabilities for each outcome are given below. (ii) Calculate the expected value of the biased die above. Expected value: Σx.p(x) Σx.p(x) = 0·22 + 0·24 + 0·75 + 0·56 + 0·9 + 0·54 = 3·21

  28. 8. This die was used in a game where the player rolls the die once and wins back the number of euro shown on the die. The game costs €3 to play. (iii) By doing the necessarycalculations, show the differencebetweenusing the biased die and a fair die to play the game. Cost €3 Expected win €3·21 3·21 – 3 = 21 c Fair die:

  29. 8. This die was used in a game where the player rolls the die once and wins back the number of euro shown on the die. The game costs €3 to play. (iii) By doing the necessarycalculations, show the differencebetweenusing the biased die and a fair die to play the game. Cost €3 Expected win €3·50 €3·50 – €3 = 50 c Therefore, in a larger number of games the average win in higher with a fair die. 50 c > 21 c

  30. 9. While shuffling a deck of cards, two cards are accidently dropped. What is the probability that the cards are: (i) a king and a queen? P(K and Q) = P(KQ) or P(QK) = = =

  31. 9. While shuffling a deck of cards, two cards are accidently dropped. What is the probability that the cards are: (ii) both red fours? P(Red 4s) = = =

  32. 9. While shuffling a deck of cards, two cards are accidently dropped. What is the probability that the cards are: (iii) both of the same suit? P(Same Suit) = P(♦♦) or P(♥♥) or P(♣♣) or P(♠♠) = = =

  33. 9. While shuffling a deck of cards, two cards are accidently dropped. What is the probability that the cards are: (iv) both black picture cards? P(Black Picture Cards) = = =

  34. 9. While shuffling a deck of cards, two cards are accidently dropped. What is the probability that the cards are: (v) one is a spade and the other is not a spade? P(Spade and not spade) = P(♠?) or P(?♠) = = = =

  35. 9. While shuffling a deck of cards, two cards are accidently dropped. What is the probability that the cards are: (vi) at least one is a diamond? P(At least 1 ♦) = P(♦♦) or P(♦?) or P(?♦) = = =

  36. 10. A team from school A and a team from school B are in the finals of a Gaelic football competition. In this competition the final round consists of a best-of-three game series. Each game continues until one team has a winning score. The team that wins two games wins the competition. Both teams are equally likely to win a game. Team A: Team B: P(Success) = P(Success) = P(Failure) = P(Failure) =

  37. 10. A team from school A and a team from school B are in the finals of a Gaelic football competition. In this competition the final round consists of a best-of-three game series. Each game continues until one team has a winning score. The team that wins two games wins the competition. Both teams are equally likely to win a game. Find the probability that team A will win the competition at the end of the first two games. (i) P(SS) = =

  38. 10. A team from school A and a team from school B are in the finals of a Gaelic football competition. In this competition the final round consists of a best-of-three game series. Each game continues until one team has a winning score. The team that wins two games wins the competition. Both teams are equally likely to win a game. Find the probability that no winner is decided until after three games have been played. (ii) No winner until 3 games Team A: P(SF) or P(FS) = = =

  39. 10. From an analysis of previous games, it is found that the probability of winning at home for both teams is . The first game is played at school A, the second at school B and the third at school A. Draw a tree diagram to represent all possible outcomes for the three matches. Include the probabilities on each branch. (iii)

  40. 10. From an analysis of previous games, it is found that the probability of winning at home for both teams is . The first game is played at school A, the second at school B and the third at school A. Use your tree diagram to investigate a claim that team A is more likely to win the competition. Clearly justify your answer with mathematical calculations. (iv) Team A win: P(AA) or P(ABA) or P(BAA) Team A win = = = =

  41. 10. From an analysis of previous games, it is found that the probability of winning at home for both teams is . The first game is played at school A, the second at school B and the third at school A. Use your tree diagram to investigate a claim that team A is more likely to win the competition. Clearly justify your answer with mathematical calculations. (iv) Team B win: P(BB) or P(BAB) or P(ABB) Team B win = = = =

  42. 10. From an analysis of previous games, it is found that the probability of winning at home for both teams is . The first game is played at school A, the second at school B and the third at school A. Use your tree diagram to investigate a claim that team A is more likely to win the competition. Clearly justify your answer with mathematical calculations. (iv) Therefore, Team A is more likely to win the competition.

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