Teaching objectives
A primary-school lab (5th and 6th grade of primaria, ages 10-12) for discovering what “probability” means, with urns of coloured balls, spinners, coins, dice, babies and ants. Every room follows the same path: the student bets first, by intuition; then runs many experiments very quickly; then sees all the possible cases, none hidden; and finally marks the ones of interest and counts them. Each student starts with 10 Ᵽ; a won bet adds 5 Ᵽ and a lost bet takes away 3 Ᵽ. There is one bet in each room (three in the last one) and it is settled with a long series of experiments (100 to 1000) that plays itself, so deciding well pays off.
There are no explanation screens: the theory comes as challenges. First the student experiments by hand a few times (the ball comes out, the coin spins, the die rolls) and then in series, with a histogram that fills up. Then all the cases are drawn (the urn with all its balls, the table of the coins, the 6 × 6 table of the two dice, the tree, the arrows of the ants) and the student taps the ones that count and writes the fraction. The line of the calculation is drawn on the histogram only after it has been calculated, and so it is compared with what came out.
What you'll learn
The lab takes the student from “what has more cases comes out more” to using probability to decide, going through the most common traps. It is organised in five blocks of rooms.
- Comparing events by counting cases and seeing that more likely does not mean certain (rooms 1-2).
- Placing probability on a scale from 0 to 1, with 0 for the impossible and 1 for the certain (rooms 3-4).
- Using the rule “favourable cases over possible cases” and knowing when it fails: when the cases are not equally likely (rooms 5, 7 and 8).
- Understanding that ½ does not mean “once in every two”: it tells what happens in the long run, not in every pair of tosses (room 6, and it comes back in almost all the others).
- Counting all the possibilities with tables and trees so that none is forgotten (rooms 9-11).
- Solving problems with a trap, in which knowing something changes the cases that are left: the babies, John’s and Ann’s children, and the ants (rooms 12-15).
- Deciding with probability: sharing the balls between two urns to score more points, and reviewing everything with new situations (rooms 16-17).
Key mathematical ideas
- More cases, more likely. In an urn with 7 red balls, 2 blue and 1 green, red comes out about 7 times in 10. With the spinners, what counts is how much room the colour takes up in total, not whether it is together or split: 2/8, 3/8 and 4/8. But more likely is not certain: a single draw can give the green ball.
- From 0 to 1. Drawing red from an urn of red balls is certain (1) and drawing blue is impossible (0). Drawing 4 balls from an urn with 3 colours always repeats a colour (1); drawing 4 red when there are only 3 is impossible (0). With 3 balls it is no longer certain: three different colours come out 3 times in 11 and a colour repeats 8 times in 11.
- Favourable cases over possible cases. In an urn of 3 red and 5 blue balls, red is 3/8 and blue is 5/8. The rule only works if all the cases are equally likely: “prize or nothing” on a wheel with one piece out of 20 is 1/20, not ½, and “red or blue” with red on 3 pieces out of 8 is 3/8. To count cases you first have to cut into equal pieces.
- ½ is not “once in every two”. With 10 coins, exactly 5 heads comes up only 1 time in 4 (63/256 ≈ 0.25), and with 100 coins, exactly 50 heads comes up about 8 batches in 100. In 2 tosses of two coins, “one head and one tail” comes up exactly 1 time in only half of the runs. What gets close to ½ is the proportion of heads when a huge number of coins is tossed. The same goes for 1/8, 1/6 or 3/4: no probability says “one in every n” in a few tries.
- Counting all the cases. Two coins give 4 cases and “one of each” takes up two (OX and XO): ¼, ½, ¼; D’Alembert’s mistake was counting it as a single case. Three coins give 8 paths: 1/8 for three heads, 3/8 for exactly two and 7/8 for at least one. Two dice give 36 cells: the 7 has 6 (1/6) and the 2 and the 12 have one each. Four coins give 16 cases and 6 with two heads: 6/16 = 3/8.
- What happened before does not count. If two boys have already been born, the next two babies have 4 equally likely cases and “a boy and a girl” is 2/4 = 1/2.
- Knowing something changes the cases. If a family with two children has at least one girl, 3 of the 4 cases are left and both girls is 1/3, not ½. If John’s older child is a boy, 2 cases are left and two boys is 1/2; if all we know about Ann is that she has at least one boy, 3 cases are left and it is 1/3.
- The ants. With three ants on a triangle there are 8 combinations of arrows and only 2 avoid a meeting (all three going the same way round): they meet with probability 6/8 = 3/4.
- Deciding. With two urns, 10 white balls and 10 black ones, the computer picks one urn at random, and the probability of white is ½ · p₁ + ½ · p₂. Sharing half and half, or all the white together, gives ½ (about 250 points in 100 games); with one white ball alone in an urn, ½ · 1 + ½ · 9/19 = 14/19 ≈ 0.74 (about 605): it is better for one urn to win always even if the other gets a bit worse.
Level: primaria, 5th and 6th grade (ages 10-12). It is the primary-school lab of the probability series: the table of cases, the tree and the rule of favourable over possible cases are the tools that the ESO and Bachillerato labs, such as “Monty Hall”, take up later. In class it works well to project room 6 or room 9 and ask the class to vote before tossing, and to let students try the sharings of room 16 in pairs. Rooms 13 and 14 invite discussion: intuition says ½ and the table says 1/3.
Room-by-room contents
Room 1 · Which colour will come out?
An urn with 10 balls: 7 red, 2 blue and 1 green. A ball is taken out, its colour is checked and it is put back. Bet: if a ball is drawn, noted and put back, 100 times over, which colour will come out most (red, blue or green)? Six balls are drawn by hand, with a histogram of colours, and the bet is settled with a series of 100 balls; afterwards the student can keep experimenting with 100 or 1000 more balls. Then the balls in the urn are counted and “7 out of 10” is written; with that challenge the line of the calculation (0.7, 0.2 and 0.1) appears on the histogram. Finally the colours are put in order from most to least likely. What has more cases is more likely, but more likely does not mean certain: a single draw can give the green ball.
Student tasks
- Bet on which colour will come out most in 100 balls.
- Count in the urn how many red balls there are and how many there are in total (7 out of 10).
- Put the colours in order from most to least likely (red, blue, green).
Room 2 · Three spinners
Three spinners with sectors of different sizes: on A red takes up 1/4 of the spinner, on B it takes up 3/8 (in both cases in a single sector) and on C red and blue alternate over 8 equal sectors, so red adds up to half. Bet: on which spinner will red come up most often (A, B, C or equally often on all three)? All three are spun together 3 times by hand, with a red counter under each, and the bet is settled with 300 spins of each spinner and a bar chart with the reds of each spinner. Then each spinner is cut into 8 equal pieces and the red ones are counted: 2, 3 and 4 out of 8; the line of the calculation (in proportion 2 : 3 : 4) is drawn on the bars after that challenge. What counts is how much room red takes up in total, not whether it is together or split: C seems to have small red bits, but together they fill half of the spinner.
Student tasks
- Bet on which spinner red will come up on most often.
- Write how many of the 8 pieces are red on each spinner (2, 3 and 4).
- Put the spinners in order from most to least likely to give red (C, B, A).
Room 3 · From 0% to 100%
Probability goes from 0% to 100%, and this room shows what the two ends mean, with no bet and no experiments: drawing balls from an urn that only holds red ones teaches nothing you cannot already see. First an illustrated idea opens (“💡 Idea”, button “Lo entiendo →”, i.e. “I get it”): something that never happens is 0%, something that always happens is 100%, and something with the same chance of happening as not happening is 50%, with coin examples (heads and tails at once, 0%; tails, 50%; heads or tails, 100%). Then an urn with 8 balls, all red: the student is asked the probability of drawing red (100%: it is certain) and of drawing blue (0%: it is impossible). Finally six events are placed on the probability thermometer, from 0% to 100%, which has five zones from “impossible” to “certain”: rolling a 7 with a die (impossible), snow in Seville in August (unlikely), rolling an even number with a die (as likely as not, 50%: 3 in 6), rolling less than 6 with a die (very likely: 5 in every 6), and the sun rising tomorrow or drawing red from the urn of red balls (certain).
Student tasks
- Choose the probability of drawing a red ball from the urn of red balls (100%: it is certain).
- Choose the probability of drawing a blue ball from that same urn (0%: it is impossible).
- Place six events on the probability thermometer, from impossible to certain.
Room 4 · Certain?
An urn with 3 red, 4 green and 5 blue balls, from which several balls are drawn at once, without putting them back (4 in the bet and 3 later; the introduction holds for the whole room). Bet: will there always be at least two balls of the same colour (yes, always; almost always but not always; or about half of the time)? Four balls are drawn by hand 3 times, and a message says whether a colour repeats; the bet is settled with 1000 draws of 4 balls, and the colour repeats in every one: probability 100%. It is certain because there are only 3 colours and the fourth ball has to repeat one. Drawing all 4 red balls, on the other hand, is impossible (0%): there are only 3 red. Then only 3 balls are drawn, another 1000 times, and the student is asked whether it is now certain that a colour repeats: it is not, because sometimes a red, a green and a blue come out (3 times in 11); a colour repeats 8 times in 11. “Almost always” is not “always”.
Student tasks
- Bet on whether with 4 balls there will always be two of the same colour.
- Choose why a colour always repeats (there are 3 colours and 4 balls are drawn).
- Choose the probability that all 4 balls are red (0%: there are only 3 red).
- Decide whether with 3 balls it is certain that a colour repeats (no: sometimes three different colours come out).
Room 5 · Favourable cases over possible cases
The room opens with an illustrated idea (“💡 Idea”, “Lo entiendo →”): probability can be worked out without experiments, by counting cases; example, a wheel with 4 equal parts and only one red shows red 1 time in 4, 25 in 100, 25%, as long as the parts are equal. Then an urn with 8 balls: 3 red and 5 blue. Bet: if 200 balls are drawn, putting each one back, what will come out more, red, blue or equally often? Six balls are drawn by hand, with a histogram, and the bet is settled with a series of 200; afterwards the student can keep experimenting with 100 or 1000 more balls. Then the student counts the balls in the urn (how many are red and how many there are in total) and writes “3 out of 8”; the red balls are no longer marked. Here the rule is stated: probability is favourable cases over possible cases, and for drawing blue it is 5 out of 8. With that challenge the lines of the calculation (3 out of 8 and 5 out of 8) appear on the histogram. The rule only works if every ball is equally easy to draw: if some were bigger or closer, it would not work.
Student tasks
- Bet on what will come out more in 200 balls: red, blue or equally often.
- Count the balls in the urn and write how many are red and how many there are in total, in two boxes “a out of b” (3 out of 8).
- Write how many blue balls there are out of how many, with the rule of favourable over possible cases (5 out of 8).
Room 6 · 50% is not “once in every two”
It opens with an illustrated idea (“💡 Idea”, “Lo entiendo →”): to estimate a probability we can do the experiment 100 times and count how many times it happened; with a coin we expect about 50 heads and 50 tails, but it will not always be so. Then ten coins are tossed at once; each toss of 10 coins is a batch, and the heads are counted. Bet: what happens more often, exactly 5 heads, any other number of heads, or both equally often? Two batches are made by hand, with a histogram of the number of heads (the bar for 5 is gold), and the bet is settled with 200 batches: exactly 5 heads comes up only 1 time in 4 (63/256 ≈ 0.25). The student is asked whether 50% means that 10 tosses give 5 heads: it does not. With 100 coins and 1000 batches, exactly 50 heads comes up only about 8 batches in 100. What does get close to 50% is the proportion of heads: a graph follows it with 1, 10, 100, 1000 and 2000 coins in a row, and the 50% line is drawn after the challenge. Finally, with 1000 tosses about 500 heads are expected, almost surely not exactly 500, but very close. This is the idea that comes back all through the lab: 50% tells what happens in the long run, not in every pair of tosses nor in every batch of 10.
Student tasks
- Bet on whether exactly 5 heads or any other number comes up more often.
- Decide whether a probability of 50% means that 10 tosses give 5 heads (no: it happens only 1 time in 4).
- Choose what gets close to 50% when a huge number of coins is tossed (the proportion of heads).
- Write how many heads are expected in 1000 tosses (about 500).
Room 7 · Two possibilities, 50%?
Some people say: “there are two possibilities, so the probability is 50%”. The room opens with an illustrated idea (“💡 Idea”, “Lo entiendo →”): counting cases only works if all the cases are equally likely to come up; the example is the crocodile (“either it eats me or it does not: 50%” is false) and the prize-or-nothing wheel cut into 20 equal pieces, only 1 of which is prize: 1 in 20, about 5 in 100 (5%). Then it is checked with that wheel with two results: prize (yellow) or nothing (grey). Bet: in 100 spins, how many prizes will come up (between 40 and 60, fewer than 15 or more than 80)? The wheel is spun 5 times by hand, with a histogram and a prize counter, and the bet is settled with 100 spins: about 5 prizes come up, not 50. Then the wheel is cut into 20 equal pieces and the student is asked how many of them would have to be prize for the prize to come up half of the time (10 out of 20); the old challenge about the fraction that is prize is gone, because the idea now explains it. The last challenge is “you either end up in jail or not: two possibilities, so 1 in 2”, which is rejected because the two cases are not equally likely. The rule of favourable over possible cases only works if the cases are equally likely.
Student tasks
- Bet on how many prizes will come up in 100 spins.
- Choose how many of the 20 pieces would have to be prize for the prize to come up half of the time (10).
- Decide whether “you either end up in jail or not, so 1 in 2” is true (no: the two cases are not equally likely).
Room 8 · The tricky spinner
A wheel with two colours, red and blue, in which red takes up 3/8 in a single sector and blue takes up 5/8. Bet: with two colours, in 200 spins, which will come up more (red, blue or equally often)? It is spun 5 times by hand, with a histogram, and the bet is settled with 200 spins: blue wins, with about 125 spins against about 75. Then the wheel is cut into 8 equal pieces and the red pieces are tapped (3); the fraction of red is 3/8, not ½, even though there are two colours; with that challenge the line of the calculation is drawn. To count cases you first have to cut into equal pieces.
Student tasks
- Bet on which colour will come up more in 200 spins.
- Mark the red pieces of the wheel cut into 8 equal pieces (3).
- Write the probability of red as a fraction (3/8).
Room 9 · Two coins
The room opens with an illustrated idea (“💡 Idea”, “Lo entiendo →”): when two things happen at once, a table helps not to forget any case; example, an ice cream with 2 flavours and 3 toppings, 2 × 3 = 6 cells. Then a gold coin and a silver coin, tossed together; O is heads and X is tails. Bet: in 300 tosses, what will come up most (two heads, two tails, one head and one tail, or all three equally often)? They are tossed 6 times by hand, with a histogram, and the bet is settled with 300 tosses: one of each comes up half of the time, and two heads and two tails a quarter each. Then the table of all the cases is shown, with gold in the rows and silver in the columns: “one head and one tail” takes up two cells (OX and XO), so its probability is 2/4 and that of two heads is 1/4. D’Alembert’s mistake is explained: he thought all three possibilities were worth ⅓ because he counted “one of each” as a single case. Finally, in 100 runs of 2 tosses, “one head and one tail” comes up exactly 1 time in only half of the runs: ½ is not “one out of every two”.
Student tasks
- Bet on what will come up most when two coins are tossed.
- Mark the cells of the table where one head and one tail come up (2).
- Write, with two boxes “a out of b” each time, how many cells count out of how many there are: one head and one tail (2 out of 4) and two heads (1 out of 4).
- Choose where D’Alembert’s mistake was (“one of each” is two cases, OX and XO).
Room 10 · Three coins
The room opens with an illustrated idea (“💡 Idea”, “Lo entiendo →”): if several things happen one after another, a tree is drawn and each path is a case; example, red or blue shirt and then short or long trousers, 2 × 2 = 4 paths. Then three coins are tossed at once. Bet: in 200 tosses, what comes up more, three heads or exactly two heads (or equally often)? They are tossed 5 times by hand, with a histogram of the number of heads, and the bet is settled with 200 tosses. Then a tree is drawn branch by branch, coin by coin: each coin splits every path in two, and there are 8 paths in all (2 × 2 × 2). The paths with exactly two heads (OOX, OXO and XOO) are tapped, the probability of three heads (1/8) and of exactly two (3/8) is written, and the line of the calculation is drawn (1/8, 3/8, 3/8, 1/8). Finally the only path with no heads (XXX) is counted and it follows that “at least one head” has probability 7/8. Three heads is 1/8 but that does not mean “one in every 8”, and “exactly two” comes up three times as often because it has 3 paths, not 1.
Student tasks
- Bet on whether three heads or exactly two heads comes up more.
- Write how many paths there are in total in the tree (8).
- Tap the paths with exactly two heads (3).
- Write the probability of three heads (1 out of 8) and of exactly two (3 out of 8).
- Write in how many paths there is no head (1) and the probability of at least one head (7 out of 8).
Room 11 · Two dice
A red die and a blue die, rolled together and added. Bet: in 500 rolls, which sum will come up most (the 2, the 7, the 12 or all equally often)? They are rolled 6 times by hand and the bet is settled with 500 rolls and a histogram of the sums from 2 to 12, which comes out triangular. Then the 6 × 6 table appears with all the ways the two dice can land, with the red die in the rows and the blue one in the columns: the cells are counted (36), those of the 7 are marked (6) and those of the 2 and the 12 (1 each), and the probability of the 7 (6/36) and of the 12 (1/36) is written. With that challenge the line of the calculation is drawn. The 7 comes up more because it can be made in 6 ways and the 12 in one. Also, 6/36 = 1/6 does not mean “one 7 in every 6 rolls”.
Student tasks
- Bet on which sum will come up most in 500 rolls.
- Write how many cells the table has (36).
- Mark the cells with sum 7 (6) and those with sum 2 and 12 (1 each).
- Write the probability of the 7 (6/36) and of the 12 (1/36).
- Choose why the 7 comes up more than the 12 (there are 6 ways to make it and only one to make 12).
Room 12 · At the hospital
At a hospital two boys have just been born, and now two more babies will be born, one after the other. Bet: in those two births, what is more likely (two boys, two girls, a boy and a girl, or all three equally often)? Five pairs of babies are watched being born by hand, with two cots that fill up, and the bet is settled with 300 pairs of births: a boy and a girl comes up half of the time and each of the other two a quarter. The table is shown with the first baby in the rows and the second in the columns: “a boy and a girl” takes up two cells (boy then girl, and girl then boy), so its probability is 2/4 = 1/2; with that challenge the line of the calculation is drawn. The student is asked whether it changes anything that two boys were born before: it does not, each birth starts from zero. And ½ does not mean “one in every two”: with few pairs it can come out more or less often.
Student tasks
- Bet on what is more likely in the next two births.
- Tap the cells of the table where a boy and a girl are born (2).
- Write the fraction of cells that are “a boy and a girl” (2/4).
- Decide whether it changes anything that two boys were born before (no: each baby is a boy or a girl with ½).
Room 13 · At least one girl
The room opens with an illustrated idea (“💡 Idea”, “Lo entiendo →”): when we know something about the result, we cross out the cases that can no longer be and count only among those left; example, a die with “it is less than 5”: 5 and 6 are crossed out and getting a 4 is 1 out of 4, 25%. Then a family has two babies and it is known that at least one is a girl. How likely is it that the other one is a girl too? Bet: 600 families are simulated and those with at least one girl are kept (about 450); in what fraction of them are both girls (about half, about 1 in 3, or all of them)? Six families are watched appearing by hand, and those with no girl do not count and fade out; the bet is settled with 600 families, on a bar that counts “k of n”. Then the 4 possible families are shown (the older baby on the left) and the one that does not count, boy and boy, is crossed out. Three are left, and in only 1 are both girls: 1/3, not ½. “At least one girl” leaves 3 cases, not 2. And 1/3 does not mean “one in every three” either.
Student tasks
- Bet on what fraction of the families with a girl have both girls.
- Cross out the family that does not count, because it has no girl.
- Write in how many of the 3 families left both are girls (1).
- Complete “the other baby is also a girl in … out of … families” (1 out of 3).
Room 14 · John’s and Ann’s children
John and Ann do not know each other and each has two children. John’s older child is a boy; Ann has at least one boy. Bet: who is more likely to have two boys (John, Ann or equally likely)? One family of each is shown 4 times, with the older child on the left, and the bet is settled with 300 families of each: John, about 150 with two boys; Ann, about 100. Then, two tables of 4 families: in John’s the ones that cannot be because the older child is not a boy are crossed out (2 are left); in Ann’s, the one with no boy (3 are left). Two boys is 1 out of 2 for John (1/2) and 1 out of 3 for Ann (1/3). It is not the same to know which of the two children is the boy (2 cases are left) as to know only that there is at least one (3 cases are left).
Student tasks
- Bet on who is more likely to have two boys.
- Cross out in John’s table the families in which the older child is not a boy (2 are left).
- Cross out in Ann’s table the family with no boy (3 are left).
- Write in each table the fraction of families with two boys (1 out of 2 and 1 out of 3).
- Choose why it is not the same (knowing which one is the boy leaves 2 cases; “at least one”, 3).
Room 15 · The three ants
Three ants, one at each corner of a triangle. Each goes to another corner, choosing at random one of the other two; if two cross on the way or arrive at the same corner, they meet. Bet: what happens more often, two ants meet, no ants meet, or both equally often? The ants are let go 5 times by hand, with an animation that shows the meeting, and the bet is settled with 200 walks: they meet in about 150, three quarters. Then the 8 combinations of arrows are shown (each ant picks one of 2 paths), the 6 in which they meet are marked and the fraction is written: 6/8 = 3/4. They only avoid each other in the two combinations in which all three go the same way round. And 3/4 does not mean “3 walks in every 4”.
Student tasks
- Bet on whether two ants meeting or no ants meeting is more frequent.
- Tap the combinations of arrows in which two ants meet (6 out of 8).
- Write the probability that they meet (6/8 = 3/4).
- Choose when no ants meet (when all three go the same way round).
Room 16 · Share out the balls
A story: on a trip with your parents you are kidnapped by cannibals, who give you 2 urns, 10 white balls and 10 black balls. The balls must be shared between the two urns, all of them and leaving neither empty; the chief picks an urn at random and his helper draws a ball at random from that urn: if it is white, you are set free; if it is black, you are eaten. There is no bet between ready-made sharings: the student starts with two empty urns, shares out the balls and tries each sharing 100 times, with a count of “free”, and a table keeps the sharings tried. The goal is to go free more than 70 times in 100; if that is not reached, hints appear, the first after 4 tries (look at what happens with an urn holding very few balls) and the second after 8 (one single white ball in urn 1 and all the others in urn 2). With half and half, or with all the white balls together, you go free about 50 times in 100; the best sharing, one white ball alone in an urn and the rest in the other, gives 1/2 · 1 + 1/2 · 9/19 = 14/19, about 74 in 100. Then three multiple-choice challenges: with one white ball alone, if that urn is picked white comes out for sure; in the other urn (9 white and 10 black) black is more likely, 10 out of 19; and “all the white together” does not beat “half and half” because one urn saves you for sure and the other never does.
Student tasks
- Share the balls between the two urns, leaving neither empty, and try each sharing 100 times.
- Look for a sharing that sets you free more than 70 times in 100 (with hints after 4 and after 8 tries).
- Choose what happens if the chief picks the urn with a single white ball (white comes out for sure).
- Choose what is more likely in the other urn, with 9 white and 10 black (black, 10 out of 19).
- Explain why “all the white together” does not beat “half and half” (one urn saves you for sure and the other never does).
Room 17 · Bet, test and count
A review with three new situations; in each one the student bets, tests, counts every case and answers. First: an urn with 4 yellow balls and 1 red one; a bet on which colour will come out most in 100 balls, the student counts the balls in the urn (how many are red and how many there are in total) and writes “1 out of 5”; then chooses the probability of drawing green, 0% because there is none (the red ball is no longer marked). Second: a die; a bet on whether in 300 rolls more even numbers, more numbers greater than 4, or the same will come up; the 6 faces are shown and the even ones (2, 4 and 6: 3 out of 6) and those greater than 4 (5 and 6: 2 out of 6) are marked. Third: 4 coins at once; a bet on how often exactly 2 heads will come up (almost always, about 3 times in 8 or never), with 200 rounds; the table of the 16 cases, all equally likely, is shown, the 6 with two heads are marked and “6 out of 16” is written (the same as 3 in every 8). At the end, a summary of what has been learned: more likely is not certain, probability goes from 0% to 100%, favourable over possible cases only if they are equally likely, count every case with tables and trees, and ½ is not “one out of every two”.
Student tasks
- Bet on which colour will come out most in the urn of 4 yellow and 1 red.
- Count the balls in the urn and write “1 out of 5” for red, and choose the probability of drawing green (0%).
- Bet on whether a die gives more even numbers or more numbers greater than 4, and mark the faces of each case (3 out of 6 and 2 out of 6).
- Bet on how often exactly 2 heads will come up with 4 coins, and mark the cases with 2 heads (6 out of 16).
- Write how many cases have exactly 2 heads and how many there are in total (6 out of 16, or 3 in every 8).
Rooms to project
The most striking ones to show and discuss in class.