Probability Calculator

Calculate probability of events with step-by-step explanations

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∑Math Input
Probability of rolling a 6 on a fair die
Probability of getting heads twice in 3 coin flips
A bag has 5 red and 3 blue balls. What is the probability of drawing a red ball?

What is Probability?

Probability measures how likely an event is to occur. It is expressed as a number between 00 and 11 (or equivalently, 0%0\% to 100%100\%).

P(A)=Number of favorable outcomesTotal number of possible outcomesP(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}

Key Concepts

  • Sample space SS: the set of all possible outcomes
  • Event AA: a subset of the sample space
  • Complement A′A': the event that AA does NOT occur; P(A′)=1−P(A)P(A') = 1 - P(A)

Types of Probability

  • Theoretical probability: Based on reasoning about equally likely outcomes (e.g., a fair coin has P(heads)=12P(\text{heads}) = \frac{1}{2})
  • Experimental probability: Based on observed frequencies from experiments
  • Subjective probability: Based on personal judgment or expertise

Probability Rules

  • 0≤P(A)≤10 \le P(A) \le 1 for any event AA
  • P(S)=1P(S) = 1 (something must happen)
  • P(∅)=0P(\emptyset) = 0 (impossible event)

How to Calculate Probability

Basic Probability

For equally likely outcomes:

P(A)=∣A∣∣S∣=favorable outcomestotal outcomesP(A) = \frac{|A|}{|S|} = \frac{\text{favorable outcomes}}{\text{total outcomes}}

Addition Rule (OR)

For the probability that event AA or event BB occurs:

P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

If AA and BB are mutually exclusive (cannot happen together):

P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B)

Multiplication Rule (AND)

For the probability that event AA and event BB both occur:

P(A∩B)=P(A)⋅P(B∣A)P(A \cap B) = P(A) \cdot P(B|A)

If AA and BB are independent:

P(A∩B)=P(A)⋅P(B)P(A \cap B) = P(A) \cdot P(B)

Conditional Probability

The probability of AA given that BB has occurred:

P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}

Binomial Probability

The probability of exactly kk successes in nn independent trials, each with probability pp:

P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

where (nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!}

Summary Table

ScenarioFormula
Single eventP(A)=favorabletotalP(A) = \frac{\text{favorable}}{\text{total}}
ComplementP(A′)=1−P(A)P(A') = 1 - P(A)
A or B (general)P(A)+P(B)−P(A∩B)P(A) + P(B) - P(A \cap B)
A and B (independent)P(A)⋅P(B)P(A) \cdot P(B)
Conditional$P(A
Binomial(nk)pk(1−p)n−k\binom{n}{k} p^k (1-p)^{n-k}

Common Mistakes to Avoid

  • Assuming events are independent when they are not — drawing cards without replacement changes probabilities after each draw.
  • Forgetting to subtract the overlap in the addition rule — when events can occur together, you must subtract P(A∩B)P(A \cap B) to avoid double-counting.
  • Confusing "and" with "or" — "and" means both events happen (multiply probabilities for independent events); "or" means at least one happens (add probabilities).
  • Not considering all possible outcomes in the sample space — make sure to count the total correctly, especially with combinations and permutations.
  • Confusing conditional probability direction — P(A∣B)P(A|B) is not the same as P(B∣A)P(B|A).

Examples

Step 1: Favorable outcomes: there are 44 kings in a deck
Step 2: Total outcomes: there are 5252 cards total
Step 3: P(king)=452=113P(\text{king}) = \frac{4}{52} = \frac{1}{13}
Answer: P(king)=113≈0.0769P(\text{king}) = \frac{1}{13} \approx 0.0769

Step 1: This is a binomial probability with n=3n=3, k=2k=2, p=0.5p=0.5
Step 2: P(X=2)=(32)(0.5)2(0.5)1=3⋅0.25⋅0.5P(X=2) = \binom{3}{2} (0.5)^2 (0.5)^1 = 3 \cdot 0.25 \cdot 0.5
Step 3: P(X=2)=3⋅0.125=0.375P(X=2) = 3 \cdot 0.125 = 0.375
Answer: P(X=2)=38=0.375P(X=2) = \frac{3}{8} = 0.375

Step 1: Probability first ball is red: P(R1)=58P(R_1) = \frac{5}{8}
Step 2: After drawing one red, probability second is red: P(R2∣R1)=47P(R_2|R_1) = \frac{4}{7}
Step 3: P(both red)=P(R1)⋅P(R2∣R1)=58⋅47=2056=514P(\text{both red}) = P(R_1) \cdot P(R_2|R_1) = \frac{5}{8} \cdot \frac{4}{7} = \frac{20}{56} = \frac{5}{14}
Answer: P(both red)=514≈0.357P(\text{both red}) = \frac{5}{14} \approx 0.357

Frequently Asked Questions

The probability of an impossible event is 0. An impossible event has no favorable outcomes in the sample space, so the ratio of favorable to total outcomes equals zero.

Independent events do not affect each other's probabilities (like flipping two coins). Mutually exclusive events cannot happen at the same time (like rolling a 3 and a 5 on one die). Mutually exclusive events with nonzero probability are never independent.

With replacement, probabilities stay the same for each draw because the item is returned. Without replacement, probabilities change after each draw because the total number of items decreases and the composition changes.

Conditional probability P(A|B) is the probability of event A occurring given that event B has already occurred. It narrows the sample space to only outcomes where B is true, then checks how many of those also satisfy A.

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