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Formule
Conditional probability
The conditioning probability must be nonzero.
Parte 1 di 1
For equally likely finite outcomes, count favourable and total cases. A conditional probability restricts the denominator to the conditioning event. Drawing without replacement changes later probabilities. Independent events allow multiplication of probabilities; disjoint alternatives allow addition. A mean weights every observation, while a median depends on ordered position.
Apri immagine a grandezza naturaleTwo-draw probability tree: first red 4/6 or blue 2/6; after red, red 3/5 or blue 2/5; after blue, red 4/5 or blue 1/5. Both-red probability is 2/5.
Original Weprepuni diagram
Esempio svolto
A bag contains 4 red and 2 blue balls. Draw two uniformly without replacement. Find P(both red).
Step 1 — count the first draw. P(first red) = 4/6. If it is red, there are 3 red balls among 5 remaining balls, so P(second red | first red) = 3/5.
Step 2 — multiply along the branch. P(both red) = (4/6)(3/5) = 12/30 = 2/5. The draws are not independent: the second probability depends on the first colour.
Step 3 — check using unordered pairs. There are 6 choose 2 = 15 equally likely pairs, of which 4 choose 2 = 6 are red-red. The same probability is 6/15 = 2/5. The other tree branches sum with red-red to 1.
For a separate conditional example, two independent fair dice have 36 equally likely ordered outcomes. “At least one die shows 6” leaves 6 + 6 − 1 = 11 outcomes: the outcome (6, 6) would otherwise be counted twice. Exactly one of those 11 has two sixes, so P(both 6 | at least one 6) = 1/11. This information is different from being told that the first die shows 6.
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