Probabilistic Reasoning curriculum

Count the outcomes that match the event, count all the possible outcomes, and divide. A box of 40 pens with 14 red ones gives a chance of 14 ÷ 40 = 0.35, or 35%, of picking red. Nothing else needs to be added or multiplied for a single pick. The common slip is dividing by the wrong number, such as the count of non-red pens.

Dividing favorable by total assumes each outcome has the same chance. A spinner with one half and two quarter sections has three sections, but landing on the half is not one chance in three; it is one in two. When outcomes differ in size or weight, measure each one's share instead of counting them.

A probability of 0 means the event cannot happen and 1 means it is certain; everything else sits between. If a calculation gives 1.3, or 130%, something was added that should have been multiplied, or a group was counted twice. Use this as a quick check on every answer.

The same chance can be written as 3/8, 0.375 or 37.5%. To turn a fraction into a percent, divide and multiply by 100: 3 ÷ 8 = 0.375, so 37.5%. Converting both chances to percents is the quickest way to compare, say, 5/12 with 40%: 5/12 is about 41.7%, so it is slightly larger.

Odds of "7 to 3 against" mean 7 ways to lose for every 3 ways to win, out of 10 in all, so the probability of winning is 3/10. Going the other way, a probability of 2/9 means 2 wins and 7 losses, odds of 7 to 2 against. Add the two sides of the odds to get the total before you divide.

An event and its complement always add up to the whole. If a flight has a 45% chance of a delay, the chance it leaves on time is 100% − 45% = 55%. Questions phrased with "not", "fails" or "is on time" usually want the complement, so read which outcome is asked for before you answer.

Counting every way to get at least one success is slow; there is usually only one way to get none. In four coin flips, the chance of no heads is 1/2 × 1/2 × 1/2 × 1/2 = 1/16. So the chance of at least one head is 1 − 1/16 = 15/16. Use this route whenever "at least one" appears.

If two independent backups each work 80% of the time, each fails 20% of the time. Both fail together with probability 0.2 × 0.2 = 0.04, so at least one works 96% of the time. Adding independent safeguards multiplies the failure chances together, which is why each extra layer helps so much.

Independent events do not affect each other, so the chance of both is the product. With a 30% chance of a sale on Monday and an independent 90% chance of a delivery that day, both happen with probability 0.3 × 0.9 = 0.27, or 27%. Adding them, to 120%, is the classic error, and it gives a number no probability can be.

A bag holds 8 red, 5 blue and 7 green counters. One draw cannot be both red and blue, so the chance of red or blue is 8/20 + 5/20 = 13/20. Adding works because the two events never overlap. Multiplying here would answer a different question, about two separate draws.

Drawing one card, the chance of a heart or a 7 is not 13/52 + 4/52, because the 7 of hearts is counted in both. Subtract it once: 13/52 + 4/52 − 1/52 = 16/52 = 4/13. Ask whether both events can occur together; if they can, the overlap must come out.

"Rainy and cold" can happen only on rainy days, so it cannot be more likely than "rainy". A detailed story can feel more believable than a plain one, because each detail seems to fit, yet every added condition can only lower the chance. Rating a combined event above one of its parts is called the conjunction fallacy.

A coin flip and a die roll are independent: getting heads leaves the die's chances untouched. Two draws from a bag without putting the first back are not independent, because the first draw changes what is left. Decide which case you are in before you multiply, because the product rule needs independence.

Writing the full set of outcomes prevents a quick miscount. Flipping a coin and spinning a three-color spinner gives 2 × 3 = 6 equal outcomes, from heads-red to tails-blue. Count the ones that match the event over all 6. Listing also shows outcomes that look alike but are different, such as heads then tails versus tails then heads.

Each die has 6 faces, and the pair has 6 × 6 = 36 outcomes, with a 5 then a 6 different from a 6 then a 5. A total of 11 comes from just those two, so its chance is 2/36 = 1/18. A total of 12 needs a double 6, so 1/36. Middle totals have more combinations than the extremes, so never treat the 11 possible totals as equally likely.

One coin has 2 outcomes, two coins 4, three coins 8, and n coins 2^n. A specific sequence, such as heads, tails, heads, is just one of those outcomes. With six flips there are 2^6 = 64 outcomes, so any one exact sequence has a chance of 1/64.

Most card questions are counts over 52. Each suit holds an ace, the numbers 2 to 10 and three picture cards. Each rank appears once per suit, so four times in the deck. Work out how many cards fit the description, then divide by 52 and simplify: the 2s and 3s together are 8 cards, so 8/52 = 2/13.

A fair eight-sided die has faces 1 to 8. The multiples of 3 among them are 3 and 6, so the chance of rolling one is 2/8 = 1/4. The method is the same for any fair die; only the total changes. List the faces that qualify rather than reasoning from a six-sided die out of habit.

A jar holds 7 red and 3 blue balls. The chance the first is red is 7/10. If it is not put back, the jar holds 6 red out of 9, so the chance both are red is 7/10 × 6/9 = 42/90 = 7/15. Recompute both the top and the bottom before you multiply.

If the first ball goes back, the jar is the same for the second draw, and the chance of two reds from 7 red and 3 blue is 7/10 × 7/10 = 49/100. Without replacement it is 7/15, a little lower, because one red is gone. Read whether the item puts the item back; the two answers are usually both among the options.

Of 80 club members, 30 play chess, and 12 of those also play go. Picking at random among the chess players, the chance of a go player is 12/30 = 40%, not 12/80 = 15%. "Given that" or "of those" tells you to shrink the denominator to the known group.

Most professional basketball players are tall, but most tall people are not professional basketball players. The two conditionals use different groups as their base, the players in one case and all tall people in the other. When a claim reverses a conditional, check which group it divides by.

When counts are given by two categories, such as grade and club membership, the condition picks out one row or column. The answer is the cell you want divided by that row's or column's total. Dividing by the grand total instead answers a different question, the chance of both together.

Picture 10,000 people. If 0.5% have a condition, that is 50 people. A test catching 98% of them flags 49. If it wrongly flags 4% of the 9,950 healthy people, that is 398 more. Counts make the next step a simple division, where percentages invite confusion.

Continuing the 10,000 people: 49 true positives and 398 false ones make 447 positives. Only 49 of those 447 have the condition, about 11%. The test's 98% describes how often it catches the condition, not how often a positive is right. Confusing the two is the most common error in these items.

False positives come from the large healthy group and true positives from the small sick group. A rare condition means a huge healthy group, so a small error rate there can outnumber every real case. That is why screening tests for rare conditions are followed by a second test. Ignoring how common the condition is, the base rate, is called base-rate neglect.

A game pays $30 on heads and costs $10 on tails. The expected value is 0.5 × $30 + 0.5 × (−$10) = $15 − $5 = $10 per play. Write each outcome with its sign, multiply by its chance, and add. A loss always enters as a negative number.

If a game has a 20% chance to win $50 and an 80% chance to lose $5, the expected value is 0.2 × $50 − 0.8 × $5 = $10 − $4 = $6. Averaging the two prizes as if they were equally likely gives the wrong answer, because the loss happens four times as often.

A game with an expected value of $6 may pay $50 or cost $5 on any one play, and never exactly $6. The expected value describes where the average settles over many plays. Use it to compare games or bets, not to predict one outcome.

A $5 ticket with a 1-in-10-million chance of $20 million is worth $20,000,000 ÷ 10,000,000 = $2 on average, so the net expected value is $2 − $5 = −$3 per ticket. A huge prize can hide a poor bet, and multiplying it by its tiny chance shows the true value.

After a fair die shows no 6 in ten rolls, the chance of a 6 on the next roll is still 1/6. The die cannot track its history. Believing an outcome is "due" because it has not happened lately is the gambler's fallacy. Ask whether the events are independent; if they are, the past changes nothing.

In 20 fair coin flips, a run of four or more of the same face happens about three times in four. Real randomness clumps, so a long run is not evidence that a coin or a process is broken. A sequence that switches back and forth with no long runs looks random to most people, yet it is the less typical result.

A fair coin flipped 10 times comes up at least 70% heads about 17% of the time; over 1,000 flips that almost never happens. As trials increase, the observed share settles toward the true chance, which is the law of large numbers. Expect the most unusual figures to come from the smallest groups, before you look for a cause.

Results mix skill and luck. An unusually high result usually had good luck in it, and luck does not repeat on demand, so the next result tends to sit closer to the average. A class that scored far above its usual level on one quiz will probably score lower on the next, even if nothing changed. Check for this before crediting a change to any cause.

Judging how likely something is by how easily examples come to mind is called the availability heuristic. Vivid or widely reported events are easy to recall, so judged this way they seem more common than their rate supports, while frequent but unremarkable events seem rarer. Ask for the count or the rate rather than trusting how familiar an event feels.

Suppose there are ten doors with one prize, you pick one, and a host who knows where the prize is opens eight empty doors. Your first pick had a 1/10 chance and still does. The other unopened door carries the remaining 9/10, so switching wins nine times in ten. The host's knowledge is what makes the two doors unequal.

Asking only the diners who came back to a restaurant whether they like it, or studying only the companies still in business to learn what makes firms last, picks cases by the very outcome being studied. The ones who stayed away or closed are missing, so the result looks better than it is. Before trusting a figure, ask who could have ended up out of the sample.