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

The probability is 1/31/3. This is calculated using Bayes' theorem where event AA is 'two girls' and event BB is 'at least one girl'. The prior probability P(A)P(A) is 1/41/4 (outcome GG). The likelihood P(B∣A)P(B \mid A) is 11 (if there are two girls, there is certainly at least one). The marginal probability P(B)P(B) is 3/43/4 (outcomes GG, GB, BG). Thus, P(A∣B)=1⋅1/43/4=1/3P(A \mid B) = \frac{1 \cdot 1/4}{3/4} = 1/3.

Conditions

  • Four equally likely ordered outcomes: GG, GB, BG, BB.
  • Event A: Two girls.
  • Event B: At least one girl.

Reasoning, step by step

  1. Identify the sample space: {GG, GB, BG, BB}.
  2. Define Event A (two girls): {GG}. P(A)=1/4P(A) = 1/4.
  3. Define Event B (at least one girl): {GG, GB, BG}. P(B)=3/4P(B) = 3/4.
  4. Calculate P(B∣A)P(B \mid A): Given two girls, the probability of at least one girl is 1.
  5. Apply Bayes' theorem: P(A∣B)=P(B∣A)P(A)P(B)P(A \mid B) = \frac{P(B \mid A)P(A)}{P(B)}.
  6. Compute the result: 1⋅1/43/4=1/3\frac{1 \cdot 1/4}{3/4} = 1/3.

Example

The script states: 'Let AA be two girls and BB be at least one girl. Then P(B∣A)=1P(B\mid A)=1, P(A)=1/4P(A)=1/4 and P(B)=3/4P(B)=3/4, giving P(A∣B)=(1⋅1/4)/(3/4)=1/3P(A\mid B)=(1\cdot1/4)/(3/4)=1/3.'

Common misconceptions

  • Assuming the probability is 1/21/2 by ignoring the order of birth or the specific composition of the sample space.
  • Confusing the conditioning event 'at least one girl' with 'the first child is a girl'.

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