Math Buddyv1.0

Probability calculator for multiple events

Type or scan a question about several events. MathBuddy combines their probabilities step by step.

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What this page covers

  • Independent events: all of them, none of them, at least one, exactly one.
  • Dependent events, such as drawing without replacement.
  • Two or three events with given probabilities as fractions, decimals or percentages.
  • Not covered here: a single event from counting outcomes (use Probability calculator) and the probability of one event given another (use Conditional probability calculator).

How to enter the problem

  • Type it or paste it. Give each probability, for example P(A)=0.5, say whether the events are independent, and say what you want to find.
  • Scan it. Fit the one problem in the frame.
  • Choose a photo. Several problems in one photo are solved as a list.
  • Then tap Solve.

Worked examples

Example 1

Problem

Three independent machines work on a given day with probabilities 0.9\displaystyle 0.9, 0.8\displaystyle 0.8 and 0.7\displaystyle 0.7. What is the probability that all three work?

Answer

Verified

The answer is

0.5040.504

Explanation

  1. Name the rule
    For independent events, the probability that all happen is the product
    P(A∩B∩C)=P(A)P(B)P(C)P(A \cap B \cap C) = P(A)P(B)P(C)
  2. Multiply
    0.9×0.8×0.7=0.5040.9 \times 0.8 \times 0.7 = 0.504

Example 2

Problem

For the same three machines, what is the probability that exactly one works?

Answer

Verified

The answer is

0.0920.092

Explanation

  1. List the three cases
    Only the first works, only the second works, or only the third works. The cases cannot happen together, so their probabilities add.
  2. Find each case
    Use 1−p1 - p for a machine that fails: 0.9×0.2×0.3=0.0540.9 \times 0.2 \times 0.3 = 0.054, 0.1×0.8×0.3=0.0240.1 \times 0.8 \times 0.3 = 0.024, 0.1×0.2×0.7=0.0140.1 \times 0.2 \times 0.7 = 0.014.
  3. Add the cases
    0.054+0.024+0.014=0.0920.054 + 0.024 + 0.014 = 0.092

Example 3

Problem

A bag holds 4 red and 6 blue counters. Three are drawn one after another without replacement. What is the probability that all three are red?

Answer

Verified

The answer is

130\frac{1}{30}

Explanation

  1. First draw
    P(red)=410P(\text{red}) = \frac{4}{10}
  2. Second draw
    One red is gone, so P(red)=39P(\text{red}) = \frac{3}{9}.
  3. Third draw
    P(red)=28P(\text{red}) = \frac{2}{8}
  4. Multiply the conditional probabilities
    410×39×28=24720=130\frac{4}{10} \times \frac{3}{9} \times \frac{2}{8} = \frac{24}{720} = \frac{1}{30}

Common mistakes

  • Adding the probabilities for at least one: 0.5+0.4+0.2=1.10.5 + 0.4 + 0.2 = 1.1 is more than 1. Use 1−P(none)1 - P(\text{none}) instead.
  • Multiplying probabilities of events that are not independent, such as draws without replacement, without updating the counts.
  • For exactly one, multiplying only the event that happens and forgetting the factors 1−p1 - p for the events that do not.
  • Treating mutually exclusive events as independent. Events that cannot happen together are not independent unless one has probability 0.

Checks, assumptions and limits

  • Check with the complement. P(at least one)+P(none)=1P(\text{at least one}) + P(\text{none}) = 1. For the runnable example, P(none)=0.5×0.6×0.8=0.24P(\text{none}) = 0.5 \times 0.6 \times 0.8 = 0.24 and 0.76+0.24=10.76 + 0.24 = 1.
  • Independence is an assumption the problem has to state or justify. Without it, the product rule needs conditional probabilities.
  • MathBuddy can make mistakes. Double check important steps.

Related tasks

Frequently asked questions

Find the probability that none of the events happen, then subtract it from 1. For independent events, none is the product of the values 1 − p.

Yes, when the problem gives the conditional probabilities or describes the draws, as in drawing without replacement. Each step shows the updated probability.

Yes. Type 50% or 0.5. Ask a follow-up question in the same thread if you want the answer in another form.

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