HexMathv1.0.3

Probability calculator

Type or scan a chance question. HexMath counts the outcomes and gives the probability as a fraction.

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What this page covers

  • Equally likely outcomes: coins, dice, cards, marbles and spinners.
  • The complement rule, for questions with not or at least one.
  • Counting with combinations, such as choosing items from a group.
  • Not covered here: chains of several events with and or or (use Multiple event probability calculator) and odds (use Odds calculator).

How to enter the problem

  • Type it or paste it. Write the question in words, with the numbers, as it appears in the problem.
  • 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

A bag holds 5 red, 3 blue and 2 green marbles. One marble is drawn at random. What is the probability that it is not blue?

Answer

Verified

The answer is

710\frac{7}{10}

Explanation

  1. Count all outcomes
    There are 5+3+2=105 + 3 + 2 = 10 marbles, each equally likely.
  2. Find the probability of the opposite event
    P(blue)=310P(\text{blue}) = \frac{3}{10}
  3. Use the complement rule
    P(not blue)=1−310=710P(\text{not blue}) = 1 - \frac{3}{10} = \frac{7}{10}

Example 2

Problem

Two fair dice are rolled. What is the probability that the sum is at least 10?

Answer

Verified

The answer is

16\frac{1}{6}

Explanation

  1. Count all outcomes
    Each die has 6 faces, so there are 6×6=366 \times 6 = 36 equally likely ordered pairs.
  2. List the favorable pairs
    Sum 10: (4,6),(5,5),(6,4)(4,6), (5,5), (6,4). Sum 11: (5,6),(6,5)(5,6), (6,5). Sum 12: (6,6)(6,6). That is 6 pairs.
  3. Divide
    P=636=16P = \frac{6}{36} = \frac{1}{6}

Example 3

Problem

A box of 10 bulbs has 3 faulty ones. You pick 3 bulbs at random. What is the probability that exactly one is faulty?

Answer

Verified

The answer is

2140\frac{21}{40}

Explanation

  1. Count all ways to pick 3 bulbs
    (103)=120\binom{10}{3} = 120
  2. Count the favorable picks
    Choose 1 of the 3 faulty bulbs and 2 of the 7 good ones
    (31)(72)=3×21=63\binom{3}{1}\binom{7}{2} = 3 \times 21 = 63
  3. Divide and simplify
    P=63120=2140P = \frac{63}{120} = \frac{21}{40}

Common mistakes

  • Dividing the number of heads by the number of flips. For 2 heads in 5 flips that gives 25\frac{2}{5}, not 516\frac{5}{16}.
  • Counting (4,6)(4,6) and (6,4)(6,4) as one outcome. With two dice the ordered pairs are the equally likely outcomes.
  • Adding probabilities that overlap, or forgetting the complement when the question says at least one.
  • Using permutations when the order of the chosen items does not matter.

Checks, assumptions and limits

  • Check the range. A probability is between 0 and 1, and the probabilities of all outcomes add up to 1. For the runnable example, the chances of 0 to 5 heads add up to 3232\frac{32}{32}.
  • This page assumes equally likely outcomes, fair coins and dice, and random draws unless the problem says otherwise.
  • HexMath can make mistakes. Double check important steps.

Related tasks

Frequently asked questions

The final answer is an exact fraction where the question allows it. Ask a follow-up question in the same thread if you want it as a decimal or a percentage.

Yes. Write the question in words with the numbers. The steps count the outcomes, then divide.

Yes. Use the camera on the one problem, or choose a photo. Several problems in one photo are solved as a list.

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Last updated: · HexMath