Part I 1. On a regular six-sided die, what is the probability that you will roll an even number if you throw the die? 1/6 1/3 1/2 5/6 2. Let's say you and your friend play a game that involves several coin flips. You flip three coins at the same time. If all three coins come up the same (all heads or all tails), your friend wins. Otherwise, you win. What is the probability that you win? 3/4 1/2 3/8 1/4 3. A spinner is divided into five equal sections, with each section having a different number from 1-5 written on it. When you spin the spinner once, the arrow lands on 1. What is the probability that the spinner lands on 1 when you spin it again? 2/5 1/25 1/5 0 4. You play a game in which you flip a coin until you get tails. Each time your coin comes up heads, you win $200. If you play the game only once, what is the probability that you make exactly $1000? 1/2 1/32 1/64 1/16 5. A spinner is divided into twelve equal sections and each section is then labeled with a different number from one to twelve. What is the probability of spinning a composite number if you spin the spinner once? 5/12 1/2 7/12 1/3 6. The probability of picking a red marble from a bag of red, blue, and yellow marbles is 7/15. The probability of picking a yellow marble is 20%. There are 15 blue marbles in the bag. How many total marbles are there in the bag? Answer: (Number) 7. To win a game, I must do three things: Flip heads on a fair coin toss, roll a composite number on a number cube, and spin a 2 on a spinner with three equal areas, each labeled with a different number 1-3. What is the probability that I win the game? 1/8 1/18 1/9 2/9 8. There are four white marbles in a bag and six silver ones. Two marbles are drawn at random, one after the other. The first one is NOT replaced. What is the probability of drawing two white marbles in a row? 2/15 1/5 4/25 4/15 9. Determine the probability that a randomly selected number between one and ten (including one and ten) is a factor of 24. 1/4 1/2 3/5 1/3 10. A couple has two children. At least one of them is male. What is the probability that one child is female, assuming the probability of having either sex is equal? 3/4 1/4 1/2 2/3 Answers 1.The correct answer was 1/2. There are six numbers on a die: 1, 2, 3, 4, 5, and 6. There are three even numbers on the die: 2, 4, and 6. 3 out of six numbers are even. 3/6 reduces to 1/2. 2. The correct answer was 3/4. A quick way of going about this problem is determining the probability that one of your friend's outcomes will appear. The probability of multiple events happening is equal to the product of the probability of each single event. The probability of flipping heads is 1/2, so the probability of flipping three heads in a row would be (1/2 * 1/2 * 1/2), or 1/8. The same thing occurs in the tails. (1/2 * 1/2 * 1/2) is equal to 1/8. The probability of your friend winning is 2/8, or 1/4. Subtract 1/4 from 1 to get 3/4, the probability of you winning 3. The correct answer was 1/5. Even though the spinner just landed on 1, it doesn't change the outcome of the next spin. There is only one way to spin a 1, and there are five spots for the spinner to land on. The probability is 1/5 4. The correct answer was 1/64. You have to flip heads five times in a row to win $1000. The probability of flipping heads once is 1/2, so the probability of flipping heads five times would be (1/2 * 1/2 * 1/2 * 1/2 * 1/2), or 1/32. However, the game is not over yet. In order to end the game, you must now flip tails. The probability of that is 1/2. 1/2 * 1/32 = 1/64. 5. The correct answer was 1/2. A composite number is a number with more than two factors. We see that 4, 6, 8, 9, 10, and 12 are composite when we look at their factors. 6 out of the 12 numbers are composite. 6/12 simplifies to 1/2. 6. The correct answer was 45 15 blue marbles has to be a fraction of the total amount of marbles. It is the fraction that the red and yellow marbles do not take up. To find the fraction, first convert 20% to a fraction. 20% is equal to 1/5. 7/15 + 1/5 (or 3/15) is equal to 10/15 or 2/3. The red and yellow marbles take up 2/3 of the bag, so we know that the blue marbles must take up the remaining third of the bag. You can now set up a formula using the variable x, where x equals the total number of marbles in the bag. 1/3 * x = 15. Multiply each side by 3 (or divide by 1/3) to get x = 45 marbles 7. The correct answer was 1/18. The probability of flipping heads is 1/2. The probability of rolling a composite number (4 and 6) is 1/3. The probability of spinning a 2 is 1/3. Multiply: 1/2 * 1/3 * 1/3 is equal to 1/18. 8. The correct answer was 2/15. On the first draw, there are four white marbles out of ten. The probability of picking one is 2/5. On the second draw, assuming we picked a white marble, there are three white marbles out of nine total marbles. The probability of picking one is 1/3. Multiply the probabilities of both outcomes together to get 2/15. 9. The correct answer was 3/5. Looking at the factors of 24, we see that 1, 2, 3, 4, 6, and 8 are the factors of 24 that are between one and ten. Six out of the ten numbers between one and ten are factors of 24. 6/10 reduces to 3/5 10. The correct answer was 2/3. Since the male child could be the first or second child, the easiest way to see this problem would be to list out all possible outcomes. The couple could have a boy, then a boy. They could also have a boy and then a girl. They could have a girl first, and then a boy. The final combination would be two girls. Since the problem states that at least one child is a boy, the final option of two girls does not work. Looking at our three remaining combinations, two of them have one female. The probability is 2/3. ---------------------------------------------------------------------------- Part II 1. A card is selected at random from a standard 52 card deck. Assuming all cards are equally likely to be selected, what is the probability that a red king was selected? 1/26 1/13 1/52 15/26 2. A card is selected at random from a standard 52 card deck. Assuming all cards are equally likely to be selected, what is the probability that a red card or a king was selected? 15/26 7/13 17/26 1/2 3. A card is selected at random from a standard 52 card deck. Assuming all cards are equally likely to be selected, what is the probability that a red card was selected given that a king was selected? 1/13 1/2 1/4 1/26 4. A card is selected at random from a standard 52 card deck. Assuming all cards are equally likely to be selected, what is the probability that a king was selected given that a red card was selected? 1/13 1/2 1/26 1/4 5. A card is selected at random from a standard 52 card deck. We are told that a king was not selected. Assuming all cards are equally likely to be selected, what is the probability that the queen of hearts was selected? 1/48 1/52 1/13 1/39 6. Suppose two cards are drawn from a standard 52 card deck without replacement (which means when we draw a card, we don't put it back). Assuming all cards are equally likely to be selected, what is the probability that the second card is an ace given that the first card is an ace? 1/17 3/52 1/221 1/13 7. Suppose two cards are drawn from a standard 52 card deck without replacement (which means when we draw a card, we don't put it back). Assuming all cards are equally likely to be selected, what is the probability that the second card is not an ace given that the first card is not an ace? 47/51 1/4 1/13 16/17 8. Suppose two cards are drawn from a standard 52 card deck without replacement (which means when we draw a card, we don't put it back). Assuming all cards are equally likely to be selected, what is the probability that both cards are red? 1/4 13/50 25/2704 25/102 9. Suppose two cards are drawn from a standard 52 card deck without replacement (which means when we draw a card, we don't put it back). Assuming all cards are equally likely to be selected, what is the probability that both cards are not aces? 188/221 15/16 144/169 12/13 10. Suppose two cards are drawn from a standard 52 card deck without replacement (which means when we draw a card, we don't put it back). Assuming all cards are equally likely to be selected, what is the probability that the second card is an ace given that the first card is a club? 2/25 3/51 1/13 4/51 Answers 1. The correct answer was 1/26. There are four kings in the deck - 2 are red and 2 are black. So the probability that a red king was selected is 2/52 = 1/26 2. The correct answer was 7/13. Of the 52 cards in the deck, 26 are red, 4 are kings, and 2 are red kings. Hence the number of cards that are red or are a king are 26 + 4 - 2 = 28. We subtract 2 since the 2 red kings are being counted twice. So the probability that a red card or a king was selected is 28/52 = 7/13. 3. The correct answer was 1/2. This is a conditional probability question. We are given that a king was selected. There are 4 kings in the deck. Of the 4 kings, 2 of them are red. So the probability that a red card was selected given that a king was selected is 2/4 = 1/2. 4. The correct answer was 1/13. This is a conditional probability question. We are given that a red card was selected. There are 26 red cards in the deck. Of the 26 red cards, 2 of them are kings. So the probability that a king was selected given that a red card was selected is 2/26 = 1/13. 5. The correct answer was 1/48. This is a conditional probability question. We are given that a king was not selected. There are 48 cards in the deck that are not kings. Of the 48 cards that aren't kings, 1 of them is the queen of hearts. So the probability that the queen of hearts was selected given that a king was not selected is 1/48. 6. The correct answer was 1/17. This is a conditional probability question. We are given that the first card is an ace. There are 51 possibilities for the second card, 3 of which are aces (we are given the first card was an ace). So the probability that the second card is an ace given that the first card is an ace is 3/51 = 1/17. 7. The correct answer was 47/51. This is a conditional probability question. We are given that the first card is an ace. There are 4 aces in the deck, so there are 48 cards that are not aces. After the first card is drawn, there are 51 cards left in the deck. Among those 51 cards, 47 are not aces. Hence the probability is 47/51. 8. The correct answer was 25/102. There are 26 red cards in the deck. So the probability that the first card is red is 26/52 = 1/2. The probability that the second card is red given that the first card is red is 25/51, since there are 25 red cards left out of the remaining 51 cards. So the probability that both cards are red is (1/2) * (25/51) = 25/102. Another solution: The sample space consists of C(52,2) = 52*51/2 = 26*51 different two card hands. The number of ways we can select 2 red cards from 26 red cards is C(26,2) = 26*25/2 = 13*25. So the probability is (13*25)/(26*51) = 25/102. 9. The correct answer was 188/221. There are 48 cards in the deck that are not aces. The probability that the first card selected is not an ace is 48/52 = 12/13. Since the first card was not an ace, there are 47 cards that are not aces among the remaining 51 cards in the deck. So the probability that the second card is not an ace given that the first card is not an ace is 47/51. Therefore, the probability that both cards are not aces is (12/13)*(47/51) = 188/221. Another solution: The sample space consists of C(52,2) = 52*51/2 = 26*51 different two card hands. There are 48 cards in the deck that are not aces, we can select two cards that are not aces from the 48 that are not aces in C(48,2) = 48*47/2 = 24*47 different ways. Hence the probability is (24*47)/(26*51) = (4*47)/(13*17) = 188/221. 10. The correct answer was 1/13 Let E be the event "the first card is a club." Let F be the event "the second card is an ace." By the formula for conditional probability, Pr(F|E) = Pr(F intersect E)/Pr(E). The probability in the denominator is the easy one: 13 of the 52 cards in the deck are clubs, so Pr(E) = 13/52. We need to compute Pr(F intersect E). Note that there are two ways the first card could be a club and the second card could be an ace: Case 1: The first card is the ace of clubs and the second card is not the ace of clubs, which has probability (1/52)*(3/51)=3/(52*51). Case 2: The first card is a club other than the ace of clubs and the second card is an ace, which has probability (12/52)*(4/51)=48/(52*51). This means the probability of F intersect E is just the sum of these probabilities, which is 51/(52*51) = 1/52. So by the formula for conditional probability, Pr(F|E)=Pr(F intersect E)/Pr(E)=(1/52)/(13/52)=1/13. ----------------------------------------------------------------------------- Part III 1. You pick two cards out of a standard pack of 52. The first card is a king. What's the probability of the second card also being a king? 1/13 1/17 3/13 1/14 2. You toss a coin four times. The first, second and third toss are heads. What's the probability of the fourth toss also being heads? 1/4 1/16 1/8 1/2 3. I choose two different numbers between 1 and 10 (1 and 10 are included). You try to guess the two numbers. How much chance do you have to guess both numbers correctly in one attempt? 1/90 1/45 1/100 1/25 4. We're playing poker with a standard pack of fifty-two cards without jokers. You have two pairs: 3, 3, 4, 4, 8. You are unaware of which cards the other players are holding. You decide to replace the 8 by another card from the pack. How do you rate your chances to turn your hand into a full house? (Full house means one trio and one pair.) 1/26 1/13 4/47 2/13 5. Ashley throws two normal, six-faced dice. How much chance does she have for a total of at least 11? 1/6 2/11 1/12 1/18 6. Eight cardboard boxes are standing on the table. Two among them contain a present, the other six are empty. You are allowed to open two boxes. How much chance do you have to find at least one present? 7/16 9/16 15/28 13/28 7. Peter, Mark, Anne and Rose are four good friends. They belong to a group of sixty pupils who are to be divided at random into three classes of twenty. What's the probability of all four friends sharing the same class? 1/12 1/81 19x18x17/59x58x57 1/64 8. I have thirty socks in total disorder in my closet. Ten are black, ten are red and ten are brown, but I can't distinguish the colours in the dark. How many socks do I have to take to have at least one pair of the same colour? 12 11 4 3 9. There are seven cups of tea. Two among them contain a deadly poison that acts within an hour. You and I both drink one cup simultaneously. How big is the chance that we both survive? 10/21 25/49 11/21 24/49 10. If someone answers these ten multiple choice questions totally at random, what is the chance that he has all answers wrong? 1/1,048,576 243/1,024 1/1,024 59,049/1,048,576 Answers 1. The correct answer was 1/17. There are fifty-one cards left. Three of those are a king, so you have three chances out of fifty-one, which equals one out of seventeen. 2. The correct answer was 1/2. The result of the first three tosses doesn't influence the fourth. The odds between heads and tails are always even. 3. The correct answer was 1/45. You have two chances out of ten to guess the first number, or 1/5. Only one of the remaining nine is correct, so you have one chance out of nine to guess that one too. 1/5 x 1/9 = 1/45. 4. The correct answer was 4/47. You need to pick a 3 or a 4. There are two 3's and two 4's left among the remaining forty-seven cards, so four cards out of forty-seven can give you a full house. It doesn't matter which five cards I have. There are 47 unknown cards to you, no matter if the 3's and 4's are in the pack or in someone else's hands. 5. The correct answer was 1/12. You can throw thirty-six possible combinations with two dice (6x6). Three of those give a sum of at least 11: 5+6; 6+5 and 6+6. The probability of throwing at least 11 is 3/36, or 1/12. 6. The correct answer was 13/28. You can calculate the probability of picking two empty boxes. If you pick one box, you have a chance of 6/8 (=3/4) to find nothing inside. Then there are seven boxes left, of which five are empty, so there's a chance of 5/7 that your second box is empty as well. 3/4 x 5/7 = 15/28 to find nothing, which means 13/28 to find at least one present. 7. The correct answer was 19x18x17/59x58x57. Mark has a 19/59 chance to be in the same class as Peter. Anne has an 18/58 chance to be in the same class as the first two, and Rose has a 17/57 chance to be in the same class as the other three 8. The correct answer was 4. There are three colours, so if I only take three socks they can all have a different colour. As soon as I take a fourth one they can't all be different. 9. The correct answer was 10/21. I have five chances out of seven to survive. If I do survive, there are six cups remaining of which two are poisoned. This would mean you have four chances out of six to survive as well(=2/3). 5/7 x 2/3 = 10/21. 10. The correct answer was 59,049/1,048,576. He has three chances out of four for each question. 3/4 x 3/4 x 3/4 x 3/4 x 3/4 x 3/4 x 3/4 x 3/4 x 3/4 x 3/4 = 59,049/1,048,576. 59,049 is the tenth power of three. 1,048,576 is the tenth power of four. ----------------------------------------------------------------------------- Part IV 1. Two six-sided dice are rolled. Find the probability the first die came up a 2 and the second die came up a 5. 1/9 1/18 1/36 1/12 2. Two six-sided dice are rolled. Find the probability that one die came up a 2 and the other came up a 6. 1/9 1/36 1/18 1/12 3. Two six-sided dice are rolled. Find the probability that at least one die is a 3. 11/36 1/6 1/2 1/3 4. Two six-sided dice are rolled. Find the probability that the sum is 8. 1/3 5/18 1/6 5/36 5. Two six-sided dice are rolled. Find the probability that the sum is 8 given that the first die was 3. 1/5 1/6 5/18 1/12 6. Two six-sided dice are rolled. Find the probability that the first die came up 3 given that the dice sum to 8. 5/18 1/6 1/5 1/12 7. Two six-sided dice are rolled. Find the probability that their sum is not 7. 3/4 5/6 2/3 23/36 8. Two six-sided dice are rolled. Find the probability that their sum is not 8, given that their sum is not 7. 23/36 3/4 5/6 2/3 9. Two six-sided dice are rolled. Find the probability that the first die is not a 5, given that the second die is not a 2. 2/3 29/36 25/36 5/6 10. Two six-sided dice are rolled. But this time, the dice aren't fair: For each die, a 1 is twice as likely to be rolled as a 2, a 2 is twice as likely to be rolled as a 3, ..., and a 5 is twice as likely to be rolled as a 6 (in other words, each number is twice as likely as the number that follows it). So what is the probability of rolling a sum of 7? 64/1023 64/1223 64/1323 64/1123 Answers 1. The correct answer was 1/36. There are 6 possibilities for the first die and 6 possibilities for the second die. Hence there are 6 * 6 = 36 possible dice rolls. There is only one of them: (2,5) which has 2 on the first die and 5 on the second die. Thus the probability is 1/36. 2. The correct answer was 1/18. This event consists of the outcomes (2,6), (6,2). Hence it has probability 2/36 = 1/18. 3. The correct answer was 11/36. The probability that the first die is 3 is 1/6, the probability that the second die is 3 is 1/6, and the probability that both dice are 3 is 1/36. By the principle of inclusion/exclusion, the probability that the first die or the second die is 3 is just 1/6 + 1/6 - 1/36 = 11/36. This probability is useful in backgammon - this is the probability of entering off the bar when there is only one open point in the opponent's table. 4. The outcomes in the event "the sum is 8" are: (2,6), (3,5), (4,4), (5,3), and (6,2). Since there are 5 ways the sum can be 8, the probability is 5/36. 5. The correct answer was 1/6. This is a conditional probability question. We are given that the first die was 3: (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) Of these 6 possibilities, (3,5) is the only outcome where the dice sum to 8. So the probability is 1/6. 6. The correct answer was 1/5. This is another conditional probability question. We are given that the dice sum to 8: (2,6), (3,5), (4,4), (5,3), (6,2) Of the 5 ways that the dice sum to 8, only one has the first die 3: (3,5). Hence the probability is 1/5. 7. The correct answer was 5/6. The ways in which a 7 can be rolled are: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). There are 6 ways a sum of 7 can be rolled. There are 36 total dice rolls, thus there are 30 ways a sum of 7 cannot be rolled. Hence the probability is 30/36 = 5/6. 8. The correct answer was 5/6. We are given that the sum is not 7. In question number 7, we saw that there were 30 ways the sum could not be 7. The ways we can roll a sum of 8 are: (2,6), (3,5), (4,4), (5,3), and (6,2). So among the 30 ways we cannot roll a sum of 7, there are 25 ways we cannot roll a sum of 8. Hence the probability is 25/30 = 5/6. 9. The correct answer was 5/6. Let E be the event "the first die is a 5" and F be the event "the second die is a 2." E and F are independent events. Hence so are the complements of E and F. This means that the probability of the complement of E given the complement of F is just the probability of the complement of E, which is 1 - 1/6 = 5/6. 10. The correct answer was 64/1323. Let x = the probability of rolling a 6. Then 2x is the probability of rolling a 5, 2*2x = 4x = the probability of rolling a 4, 8x = the probability of rolling a 3, 16x = the probability of rolling a 2, and 32x = the probability of rolling a 1. These are all the outcomes, so these probabilities must sum to 1: x + 2x + 4x + 8x + 16x + 32x = 1 which simplifies to 63x = 1 which means x = 1/63. So Pr(1) = 32/63, Pr(2) = 16/63, Pr(3) = 8/63, Pr(4) = 4/63, Pr(5) = 2/63, and Pr(6) = 1/63. There are six ways to roll a sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) By independence, their corresponding probabilities are (32/63)*(1/63), (16/63)*(2/63), (8/63)*(4/63), (4/63)*(8/63), (2/63)*(16/63), and (1/63)*(32/63). Each probability is 32/63^2. We add up these six probabilities to get the probability of rolling a sum of 7: 6*(32/63^2) = 64/1323. ----------------------------------------------------------------------------- Part V 1. There is a carnival game that requires you to roll two standard 6-sided dice. If the sum of the dice rolls is a prime number, then you win. What is the probability of winning? 5/12 17/36 4/9 1/2 2. There is a standard 52-card deck. If you draw three cards randomly, what is the probability that all three cards are spades? 23 out of 1,988 39 out of 140,608 11 out of 850 1 out of 64 3. A box contains one of each of the bills: $1, $5, $10, $20, $50, and $100. If you randomly draw three bills, then what are the odds that the three bills add to $75? 1 in 120 1 in 60 1 in 20 1 in 30 4. You are running from 3 police officers and there are 5 possible caves to hide in. You hide in a random cave before the officers see you. Each of the officers chooses a random cave to search. More than one officer can search the same cave. If there is a 100% chance a cop will find you if he searches your cave, then what are the odds you will be found? 61 out of 125 11 out of 25 3 out of 5 47 out of 75 5. You randomly draw 3 cards from a standard 52-card deck. What are the odds that you draw 3 different suits? 169/425 13/25 3/8 109/331 6. A casino game costs $5 to play where you roll two dice. If the sum of the dice rolls is 7, you win $21. If the sum of the dice rolls is not 7, then you lose and get nothing. If you keep playing and stop when you win, what are the odds you will have made money? 3750/7776 4651/7776 3125/7776 4105/7776 7. There is a box containing blue, red, yellow, and green pencils. You randomly draw one of the 50 pencils. You are 2.3 times more likely to draw a green pencil than you are to draw a blue pencil. There are 8 more green pencils than there are red pencils. If you have a 4% chance of drawing a yellow pencil, then what are the odds of drawing a red pencil? 7/25 1/5 13/50 3/10 8. You draw two random cards from a standard 52-card deck. What is the probability that you will draw at least one ace? 33/221 29/221 2/13 3/26 9. You randomly flip four coins. What is the probability that you will get 2 heads and 2 tails? 3/16 1/16 3/8 1/8 10. You randomly roll two dice. What is the probability that the product of the two numbers rolled will equal a square number? 1/9 2/9 7/36 5/18 Answers 1. The correct answer was 5/12. There are 36 possible results from rolling two dice. The sum of the dice rolls will give you a number from 2 to 12. The prime numbers between them are 2, 3, 5, 7, and 11. The possible dice rolls to get those numbers are: 1-1, 1-2, 2-1, 1-4, 4-1, 2-3, 3-2, 1-6, 6-1, 2-5, 5-2, 4-3, 3-4, 6-5, and 5-6. Since there are 15 results that win the game, the odds of winning are 15/36, or 5/12. 2. The correct answer was 11 out of 850. The probability of drawing one spade is 13/52 or 1/4. The odds of drawing another spade is 12/51. The odds of drawing a third spade is 11/50. If you multiply (1/4 * 12/51 * 11/50), you get 132/10200. This simplifies to 11/850. 3. The correct answer was 1 in 20. You must draw $20, $5, and $50 since they are the only three bills that total $75. The odds of drawing one of the needed bills is 1/2. The odds of drawing another needed bill is 2/5. The odds of drawing the last needed bill is 1/4. When you multiply the three fractions, you get 2/40 or 1/20. 4. The correct answer was 61 out of 125. First I'll figure out the odds that you will not be found. The probability of one cop not finding you is 4/5. The probability of three cops not finding you would be 4/5 * 4/5 * 4/5 = 64/125. This means that 61/125 is the probability that you will be found. 5. The correct answer was 169/425. There are 4 sets of 13 cards of the same suit, so there is a 39/51 chance that the second card will be a different suit. The probability of the third card being a different suit than the others is only 26/50 or 13/25. When you multiply 39/51 * 26/50 you get 1014/2550, or 169/425. 6. The correct answer was 3750/7776. There are 36 possible results from rolling two dice. Of those results, there are 6 that will give you a 7: 1-6, 6-1, 2-5, 5-2, 4-3, and 3-4. This means there is a 1/6 chance you will roll a 7. To figure the odds of making money, first I'll figure the odds of losing money. You have to lose 4 times in order to lose money. The odds of losing once are 5/6. If you do 5/6 to the 4th power, you get the odds of losing money, which are 4026/7776. This means the odds of making money are 3750/7776. 7. The correct answer was 3/10. I will use the first letters of the colors as variables representing the number of pencils of a certain color. Since you are 2.3 times more likely to draw a green pencil than a blue one, there are 2.3 times more green pencils than blue ones. Since there is a 4% chance of drawing a yellow pencil, there are 2 yellow pencils. The following equations are true: B + R + Y + G = 50 B = G/(2.3) R = G - 8 Y = 2 We can replace the B and R with equal variables in terms of G. G/(2.3) + G - 8 + 2 + G = 50 Simplified, we get: G/(2.3) + G + G = 56 Multiply both sides by 2.3 and we get: G + (2.3)G + (2.3)G = 128.8 Simplified again, we get: (5.6)G = 128.8 G = 23 Since there are 8 more green pencils than red ones, this means there are 15 red pencils. The odds of drawing a red pencil are 15/50 or 3/10. 8. The correct answer was 33/221. First I'll figure out the odds that you won't draw an ace. Since there are 4 aces, the odds of not drawing an ace are 48/52 or 12/13. The odds of the second card not being an ace are 47/51. Multiply 47/51 by 12/13 and you get 564/663 or 188/221. Since 188/221 are the odds you won't get an ace, the odds you will get at least 1 ace are 33/221. 9. The correct answer was 3/8. There are two possible results of flipping a coin, and 4 coin flips, so the number of possible flip results is 2 to the 4th power or 16. The possible combinations of ways to get two heads and two tails, are: HHTT, HTHT, HTTH, THHT, THTH, and TTHH. Since there are 6 results that are two heads and two tails, the probability of getting that result is 6/16 or 3/8. 10. The correct answer was 2/9. The product of the two numbers rolled will be from 1 to 36. The square numbers between them are 1, 4, 9, 16, 25, and 36. 1-1, 1-4, 4-1, 2-2, 3-3, 4-4, 5-5, and 6-6 are the 8 results you can get that will give you two numbers that will multiply to a square number. There are 36 possible results from rolling two dice, so the odds are 8/36 or 2/9. -----------------------------------------------------------------------------