Independent Events
Aligned to S-CP.A.2 — Common Core State Standards for Mathematics.
What this lesson teaches
Two events are independent when one occurring does not change the probability of the other. Test independence by checking whether P(A and B) = P(A) × P(B).
Worked examples
Example 1
Flip a coin with P(heads) = 1/2 and roll a die with P(six) = 1/6. Here P(heads and six) = 1/2 × 1/6 = 1/12, which equals the product of the probabilities, so the two events are independent.
Example 2
Draw two cards WITHOUT replacing the first. P(first is an ace) = 4/52 = 1/13. P(second is an ace | first was an ace) = 3/51, which is not 4/52. Since the conditional probability changed, the two draws are NOT independent. Replace the first card and shuffle, and they become independent again.
Example 3
Common mistake: confusing independent with mutually exclusive. Rolling a 2 and rolling a 5 on one die are mutually exclusive — they cannot both happen, so P(both) = 0. But that makes them strongly DEPENDENT: knowing one occurred tells you the other did not. Independent means P(A and B) = P(A) × P(B); mutually exclusive means P(A and B) = 0. Two events with nonzero probability cannot be both.
Practice questions
- Are two separate coin flips independent?
- If P(A) = 0.4, P(B) = 0.5, and A and B are independent, find P(A and B).
- P(A) = 0.3, P(B) = 0.6, and P(A and B) = 0.2. Are A and B independent?
Watch the lesson
Every lesson comes with a video taught in English and Spanish — the same video the QR code in the printed workbook opens.
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Eventos independientes
Dos eventos son independientes cuando el que ocurra uno no cambia la probabilidad del otro. Prueba la independencia comprobando si P(A y B) = P(A) × P(B).
Ejemplo 1: Lanza una moneda con P(cara) = 1/2 y un dado con P(seis) = 1/6. Aquí P(cara y seis) = 1/2 × 1/6 = 1/12, que es igual al producto de las probabilidades, así que los dos eventos son independientes.
Ejemplo 2: Saca dos cartas SIN reponer la primera. P(la primera es un as) = 4/52 = 1/13. P(la segunda es un as | la primera fue un as) = 3/51, que no es 4/52. Como la probabilidad condicional cambió, los dos sorteos NO son independientes. Repón la primera carta y baraja, y vuelven a ser independientes.
Ejemplo 3: Error común: confundir independiente con mutuamente excluyente. Sacar un 2 y sacar un 5 en un mismo dado son mutuamente excluyentes — no pueden ocurrir a la vez, así que P(ambos) = 0. Pero eso los hace fuertemente DEPENDIENTES: saber que uno ocurrió te dice que el otro no. Independiente significa P(A y B) = P(A) × P(B); mutuamente excluyente significa P(A y B) = 0. Dos eventos con probabilidad no nula no pueden ser ambas cosas.
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