Class 12 Probability · Case Study4 Marksbayes_theorem
Class 12 Maths · Probability · Case Study #131
In a hypothetical handedness model, parental categories A,B,C,D each have probability1/4: A=both parents left-handed; B=father right-handed,mother left-handed; C=father left-handed,mother right-handed; D=both right-handed. For L=“child left-handed”, the conditional probabilities P(L|A),P(L|B),P(L|C),P(L|D) are0.24,0.22,0.17,0.09 respectively. These are given modelling assumptions.
Part (1) 1 Marks
Find P(L|C).
Answer: 17/100
- Read the conditional probability assigned to category C.
- P(L|C)=0.17=17/100.
Part (2) 1 Marks
Find P(L′|A).
Answer: 19/25=0.76
- Conditional complements satisfy P(L′|A)=1−P(L|A).
- Thus P(L′|A)=1−0.24=0.76=19/25.
Part (3) 2 Marks
Find P(A|L).
Answer: 1/3
- By total probability P(L)=(1/4)(0.24+0.22+0.17+0.09)=0.18.
- The joint probability P(A∩L)=(1/4)(0.24)=0.06.
- Bayes gives P(A|L)=0.06/0.18=1/3.
Part (4) 2 Marks
Find P(L|exactly one parent is left-handed).
Answer: 39/200=0.195
- Exactly one left-handed parent is the disjoint union B∪C; its probability is1/2.
- P(L∩(B∪C))=(1/4)(0.22)+(1/4)(0.17)=0.0975.
- Thus P(L|B∪C)=0.0975/0.5=0.195=39/200, the equal-weight average of0.22 and0.17.