Lemma

Probability · Chapter 1 of 10

Counting & chance

12 curated · 4,878 from open sources

Before you startAI-drafted · not yet reviewed by an editor

Count all equally likely outcomes and count the outcomes that satisfy the condition. The probability is the number of favorable outcomes divided by the total number of outcomes: \(P(E)=\frac{\text{favorable outcomes}}{\text{total outcomes}}\). For example, rolling two dice gives \(6\times6=36\) equally likely pairs. The pairs summing to 7 are \((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\), six favorable outcomes, so \(P(\text{sum }7)=\frac{6}{36}=\frac16\). For "at least one" events, it is often easier to count the complement: outcomes where the event does not happen, then subtract from the total.

Curated

12
#TitleDifficultySourceTags

From open sources

4,878

Showing everything, including problems with no answer to check, a missing figure, or a near-copy elsewhere. Show only complete problems

#TitleDifficultySource
Page 1 of 98Next →