Math & logic · ENGLISH
Count the outcomes before you trust the odds.
Coins, counters and one small change that makes yesterday’s fraction wrong.
Two colour names are not two equal chances
Imagine a bag holding two red counters and three blue counters. Each individual counter is equally likely to be picked. There are two colour names, but five possible counters. The probability of red is 2/5, not 1/2. The question states what is equally likely; your counting must use those same objects.
Write a small sample space
For two independent fair coin tosses, write HH, HT, TH and TT. Each ordered outcome has probability 1/4. Exactly one head appears in HT and TH, so the answer is 2/4 = 1/2. Grouping everything into “two heads, one head, no heads” hides that the middle group contains two outcomes. The three labels are not equally likely.
One word changes the count
Compare exactly one head with at least one head. Exactly one includes HT and TH. At least one also includes HH, giving 3/4. Before calculating, turn the instruction into a list of qualifying results. This is especially helpful when a familiar-looking question replaces just one word.
Check whether the setup changed
Return to the bag. Remove one red counter and leave it out. One red and three blue counters remain, so the next draw has red probability 1/4. Put the removed counter back before drawing, and the original 2/5 applies again. “With replacement” is not decoration: it says whether the contents reset.
A streak does not rewrite an independent toss
If a fair coin’s tosses are independent, five previous heads do not make tails owed on the next toss. Its two possible results still each have probability 1/2. This statement relies on the model given in the question. It is not a way to prove that a real coin or real procedure is fair merely by calling it fair.
Use the method in a timed question
First identify the experiment. Then check equal likelihood, independence and replacement where relevant. Finally count the results matching the exact instruction. Those checks can be brief, but skipping one can produce a beautifully simplified fraction for the wrong problem. Practice without a clock until you can explain which outcomes your numerator counts.
The experiment determines the sample space
Two independent fair coin tosses give four equally likely ordered results: HH, HT, TH and TT. Exactly one head occurs in HT or TH, so its probability is 2/4. “At least one head” also includes HH and therefore has probability 3/4. A small change in wording changes the event.
Knowing that the first toss was heads leaves HH and HT as the relevant possibilities for the second toss. With independent fair tosses, a previous streak does not change the next toss’s probability of heads. Independence is an assumption about the experiment, not a promise that every real process behaves like a coin.
Replacement changes the second draw
A bag contains two red and three blue counters. Draw a counter uniformly at random and suppose it is red. If it is kept out of the bag, one red and three blue remain, so the next red probability is 1/4. If it is replaced and the bag mixed again, the next red probability is 2/5.
Notice what is given: the first draw was red. Without that condition, a question about the second draw requires considering both possible first colours. Write down what remains after the stated observation before calculating.
Sources & further reading
Sources support the factual background. The worked examples and game connections are written for 3 SECONDS.