Law of total probability

Theorem taught in Random experiments and probability, part of Probability

Law of total probability

The three factories as areas. From Random experiments and probability.
Type Theorem
Taught in Random experiments and probability
Part Probability
Worked examples 3
Questions using it 1

If B_1,\ldots,B_n are disjoint and their union is \Omega, then for every event A

\mathbb{P}(A)=\sum_{i=1}^n \mathbb{P}(A\mid B_i)\,\mathbb{P}(B_i).

The events A\cap B_i are disjoint and exhaust A, so the sum rule applies, and each term is the product rule.

Examples

Three factories

A chip is made in factory 1, 2 or 3 with probabilities 0.15, 0.35 and 0.50, and the defective rates are 0.01, 0.05 and 0.02.

Question. A chip in hand is defective. What is the probability that it came from factory 1?

Solution. By Bayes’ rule, the probability it came from factory 1 is

\frac{0.15\times 0.01}{0.15\times 0.01+0.35\times 0.05+0.50\times 0.02}=\frac{0.0015}{0.029}\approx 0.052.

The rare factory is unlikely to be the source, because the two larger factories produce more defectives in total even though each chip from factory 1 is less often defective.

The same calculation gives 0.0175/0.029=0.603 for factory 2 and 0.010/0.029=0.345 for factory 3. The denominator, 0.029, is the overall defective rate, from the law of total probability.

A channel with unequal inputs

A binary channel sends a 1 with probability 1/4 and a 0 with probability 3/4. A sent 1 is received as 1 with probability 0.90, and a sent 0 is received as 0 with probability 0.95.

  1. Find the probability that a 1 is received.

  2. Given that a 1 is received, find the probability that a 1 was sent.

Let S be the symbol sent and R the symbol received.

  1. By the law of total probability,

\mathbb{P}(R=1)=\mathbb{P}(R=1\mid S=1)\mathbb{P}(S=1)+\mathbb{P}(R=1\mid S=0)\mathbb{P}(S=0)=0.90\times\tfrac14+0.05\times\tfrac34=0.225+0.0375=0.2625.

  1. By Bayes’ rule,

\mathbb{P}(S=1\mid R=1)=\frac{0.225}{0.2625}=\frac67=0.857.

One received 1 in seven is a corrupted 0, because 0s are sent three times as often.

A zero received

Question page: A zero received

Zeros and ones are sent with equal probability. A sent 0 is received as 0 with probability 0.95, and a sent 1 is received as 1 with probability 0.99. A 0 is received. Find the probability that a 1 was sent.

Let B be the event that a 1 was sent and A the event that a 0 was received. Then \mathbb{P}(A\mid B)=0.01, \mathbb{P}(A\mid B^c)=0.95 and both inputs have probability 1/2, so

\mathbb{P}(B\mid A)=\frac{0.01\times 1/2}{0.01\times 1/2+0.95\times 1/2}=\frac{0.01}{0.96}=\frac{1}{96}.

Used in

See also

Product rule · Sample space · Event · Probability measure · Properties of a probability · Equally likely outcomes · Four ways of counting draws · Conditional probability

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