The outcome explorer enumerates twelve ordered pairs without replacement and sixteen with replacement. It distinguishes the overall second-red probability 1/2 from the first-red conditional probability 1/3.
Use the worked example
Choose First red and inspect the six eligible pairs: only two end in red. Enter 1/3 and Check. Then enable replacement: eight pairs remain under the same first-color condition, four favorable, giving 1/2. A specific R1 condition displays its remaining-ID pool separately. Equivalent fractions are compared as exact whole-number ratios, including long digit strings; a zero denominator is rejected.
Explain the evidence for the answer
The eligible table changes when you condition on a first draw or return the card. These are the replacement and conditional-event ideas explained in OpenStax's discussion. State the question first so its denominator is clear before counting favorable outcomes.
Use the four supplied identities: R1 and R2 represent red cards, while B1 and B2 represent blue cards. The table lists IDs so two red cards remain distinguishable. Choosing First R1 narrows the first column to that specific card rather than removing every red card.
The initial rule is two ordered draws without replacement. Each first ID is equally likely among four, and each second ID equally likely among the three remaining. Order matters: R1 then B1 differs from B1 then R1.
Remove the selected ID, not its entire color. After R1, the remaining pool is R2, B1, B2. R2 is still a red card eligible for the second draw.
R1 followed by R1 is impossible under this rule because that exact card was not returned. A repeated color, such as R1 then R2, remains possible.
The overall second-red question does not specify the first card or color. Its denominator includes all twelve eligible pairs. Do not change that total into the three-card pool for a particular known first draw unless the question supplies that condition.
Given First red, each of the two possible red first cards leaves one red ID among three remaining cards. The displayed table therefore has six rows: R1 or R2 first, each followed by three possibilities. Two rows end red, so 2/6 and 1/3 describe the same conditional answer.
First Blue leaves both red IDs among three, giving 2/3. The first draw affects the second draw's conditional probability under no replacement.
Compare 1/3 after First red with 2/3 after First blue. Each fraction belongs to its own condition. For the Overall question, return to all twelve pairs and count six red endings. Adding the two conditional fractions would combine different questions rather than count that overall set.
In the separate replacement mode, return the first card before the second draw and again select uniformly among all four IDs. There are 4 × 4 = 16 ordered pairs, including R1 then R1.
After a first R1, the second pool is R1, R2, B1, B2. Two of four are Red, so second Red is 2/4 = 1/2. The same second-red probability holds after a first Blue because the complete pool was restored.
Overall, eight of sixteen pairs end Red, again 1/2. The same marginal probability across the two modes does not mean their conditional models are identical. Given first Red, no replacement gives 1/3 while replacement gives 1/2.
Copy a prompt for this example
Build a single HTML probability explorer for four equally likely labeled cards: red R1 and R2, blue B1 and B2. Enumerate ordered pairs for two draws and show each pair's first and second ID in a readable table. Include buttons for Overall, First red, First blue, First R1, With replacement, and Without replacement.
Start with the overall question without replacement. There are twelve ordered pairs and six end red, giving 6/12. First red leaves six eligible pairs with two red endings, giving 2/6 or 1/3. First blue leaves six pairs with four red endings. First R1 permits second IDs R2, B1, B2 and gives 1/3.
With replacement, restore the first card before the second draw. There are sixteen overall pairs and eight red endings. The first-red subset contains eight pairs with four favorable endings. Keep the currently selected condition beside its denominator so a first-ID condition cannot silently change an overall result.
Add a fraction field and Check button. Accept equivalent fractions by comparing their values against the eligible pair counts. Reject blank input, malformed fractions, and a zero denominator. Changing a rule or condition must clear the old answer and feedback.
Reset returns to Overall, no replacement, and a blank answer. Use text IDs and color names, visible focus, and a small-screen table that wraps. Return one self-contained HTML file showing theoretical outcomes rather than simulated draws.
Bring the labeled cards and the eligible-pair table to the next probability discussion. NOS can preview the HTML explorer while you check the denominator for each question. Ask learners to state the replacement rule and condition before giving the fraction; these are theoretical outcomes rather than recorded draws.