Probability question 2

Question

A person has two children. One is a boy born on a Tuesday. What is the probability that the other child is also a boy?

Answer

Let event A and event B be the births of the two children. Let us assume that the probability of a boy being born is equal to the probability of a girl being born (both 50%). The possible relevant outcomes are:

Event A – Boy born on a Tuesday (probability = (1/2)x(1/7) = 1/14)

Event A – Boy born on any day except Tuesday (probability = (1/2)x(6/7) = 6/14 = 3/7)

Event A – Girl born on any day (probability = 1/2)

and

Event B – Boy born on a Tuesday (probability = (1/2)x(1/7) = 1/14)

Event B – Boy born on any day except Tuesday (probability = (1/2)x(6/7) = 6/14 = 3/7)

Event B – Girl born on any day (probability = 1/2)

(Check: 1/2 + 3/7 + 1/14 = 1)

Independently of whether event A occurs before, after or simultaneously to event B, the only relevant combinations of these outcomes are:

  • Event A – Boy born on a Tuesday (probability = (1/2)x(1/7) = 1/14) AND Event B – Boy born on a Tuesday (probability = (1/2)x(1/7) = 1/14)
  • Event A – Boy born on a Tuesday (probability = (1/2)x(1/7) = 1/14) AND Event B – Boy born on any day except Tuesday (probability = (1/2)x(6/7) = 6/14 = 3/7)
  • Event A – Boy born on a Tuesday (probability = (1/2)x(1/7) = 1/14) AND Event B – Girl born on any day (probability = 1/2)
  • Event A – Boy born on any day except Tuesday (probability = (1/2)x(6/7) = 6/14 = 3/7) AND Event B – Boy born on a Tuesday (probability = (1/2)x(1/7) = 1/14)
  • Event A – Girl born on any day (probability = 1/2) AND Event B – Boy born on a Tuesday (probability = (1/2)x(1/7) = 1/14

 

The probabilities of each of these combinations are:

Probability of combination 1) = (1/14)(1/14) = 1/196

Probability of combination 2) = (1/14)(3/7) = 3/98

Probability of combination 3) = (1/14)(1/2) = 1/28

Probability of combination 4) = (3/7)(1/14) = 3/98

Probability of combination 5) = (1/2)(1/14) = 1/28

Combinations 1), 2) and 4) are formed by two boys, hence the probability of the second child being a boy is:

((1/196)+(3/98)+(3/98)) / ((1/196)+(3/98)+(1/28)+(3/98)+(1/28)) = (13/196) / (27/196) = 13/27

Probability question 1

Question

A person has two children. One is a boy. What is the probability that the other child is also a boy?

Answer

Let event A and event B be the births of the two children. Let us assume that the probability of a boy being born is equal to the probability of a girl being born (both 50%). The possible outcomes are:

Event A – Boy (50% probability)

Event A – Girl (50% probability)

and

Event B – Boy (50% probability)

Event B – Girl (50% probability)

Independently of whether event A occurs before, after or simultaneously to event B, the only possible combinations of these outcomes are:

  • Event A – Boy AND Event B – Boy
  • Event A – Boy AND Event B – Girl
  • Event A – Girl AND Event B – Boy
  • Event A – Girl AND Event B – Girl

Therefore, if we are told that one of the children is a boy, we can have any of the first three outcomes, but not the last outcome. In only one of the equally probably first three outcomes we have two boys, hence the probability of the other child being a boy in 1/3.