Probability & statistics / English

Bayes' Theorem - The Simplest Case

Dr. Trefor Bazett · YouTube · 5:31

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YouTube · Dr. Trefor BazettOriginal video

IN THIS VIDEO

01Start from conditional probability
02Rearrange the shared intersection
03Two children, at least one girl

SELECTED MOMENT

Start from conditional probability

SUGGESTED PREPARATION

Events, intersections and conditional probability

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The explanation, unpacked.

Reviewed learning material · Video analysis · English

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Bayes' theorem follows from writing the same intersection probability in two conditional orders. The lesson derives the formula and checks it with a two-child probability example.

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Chapters

0:50Start from conditional probability 1:38Rearrange the shared intersection 3:04Two children, at least one girl 5:14Bayes' theorem gives one third

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

For $P(B)>0$ , conditional probability is $P(A\mid B)=P(A\cap B)/P(B)$ . Reversing the roles gives the corresponding expression for $P(B\mid A)$ when $P(A)>0$ .

Since $A\cap B=B\cap A$ , we have $P(A\cap B)=P(B\mid A)P(A)$ . Substitution gives $P(A\mid B)=P(B\mid A)P(A)/P(B)$ . The denominator normalizes the probability; it is not generally a prior-odds term.

The example assumes four equally likely ordered outcomes: $GG,GB,BG,BB$ . Conditioning on at least one girl removes $BB$ , leaving three possibilities. This conclusion depends on that stated conditioning event, not on a different way of selecting or reporting a child.

Let $A$ be two girls and $B$ be at least one girl. Then $P(B\mid A)=1$ , $P(A)=1/4$ and $P(B)=3/4$ , giving $P(A\mid B)=(1\cdot1/4)/(3/4)=1/3$ .

Knowledge cards

Start from conditional probability

For $P(B)>0$ , conditional probability is $P(A\mid B)=P(A\cap B)/P(B)$ . Reversing the roles gives the corresponding expression for $P(B\mid A)$ when $P(A)>0$ .

Rearrange the shared intersection

Two children, at least one girl

Bayes' theorem gives one third

Let $A$ be two girls and $B$ be at least one girl. Then $P(B\mid A)=1$ , $P(A)=1/4$ and $P(B)=3/4$ , giving $P(A\mid B)=(1\cdot1/4)/(3/4)=1/3$ .

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Bayes theorem ExplanationAt 0:50
Why this connection?
For $P(B)>0$ , conditional probability is $P(A\mid B)=P(A\cap B)/P(B)$ . Reversing the roles gives the corresponding expression for $P(B\mid A)$ when $P(A)>0$ . Since $A\cap B=B\cap A$ , we have $P(A\cap B)=P(B\mid A)P(A)$ . Substitution gives $P(A\mid B)=P(B\mid A)P(A)/P(B)$ . The denominator normalizes the probability; it is not generally a prior-odds term. The example assumes four equally likely ordered outcomes: $GG,GB,BG,BB$ . Conditioning on at least one girl removes $BB$ , leaving three possibilities. This conclusion depends on that stated conditioning event, not on a different way of selecting or reporting a child. Let $A$ be two girls and $B$ be at least one girl. Then $P(B\mid A)=1$ , $P(A)=1/4$ and $P(B)=3/4$ , giving $P(A\mid B)=(1\cdot1/4)/(3/4)=1/3$ .

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Bayes' Theorem - The Simplest Case

The explanation, unpacked.

Chapters

Learning script

Knowledge cards

Start from conditional probability

Rearrange the shared intersection

Two children, at least one girl

Bayes' theorem gives one third

Explore the knowledge in this video

Questions this video answers

What is the probability of having two girls given at least one girl in the two-child example?+

Why is the denominator in Bayes' theorem considered a normalizing constant rather than a prior-odds term?+

How is Bayes' theorem derived from the definition of conditional probability?+

How does the conditioning event affect the sample space in the two-child problem?+

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