Explanation → ProbabilityEvidenceConnection evidence is not yet available in this language.
Explanation → ProbabilityEvidenceWork through a complete Khan Academy dice problem: on a fair six-sided die, the even faces 2, 4 and 6 give a single-roll probability of 3/6=1/2. Assuming three mutually independent rolls, multiply three one-half factors to obtain 1/8 for an even result on every roll. Editorial notes distinguish fairness from independence and clarify that pairwise independence alone does not justify a three-event product.
Content location → IndependenceEvidenceFor two events, independence means one event does not change the other event probability. Repeated rolls in this example are assumed independent; a fair die alone does not imply this.
Content location → IndependenceEvidenceConnection evidence is not yet available in this language.
Explanation → Die rolling probability with independent events | Precalculus | Khan AcademyEvidenceWork through a complete Khan Academy dice problem: on a fair six-sided die, the even faces 2, 4 and 6 give a single-roll probability of 3/6=1/2. Assuming three mutually independent rolls, multiply three one-half factors to obtain 1/8 for an even result on every roll. Editorial notes distinguish fairness from independence and clarify that pairwise independence alone does not justify a three-event product.
Explanation → IndependenceEvidenceFor two events, independence means one event does not change the other event probability. Repeated rolls in this example are assumed independent; a fair die alone does not imply this.
Explanation → IndependenceEvidenceConnection evidence is not yet available in this language.
Explanation → IndependenceEvidenceFor two events, independence means one event does not change the other event probability. Repeated rolls in this example are assumed independent; a fair die alone does not imply this.
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