Three contestants aim at a target in a throwing competition. If the probability that the first contestant hits the target is 1/3 and the probability that the second hits the target is 1/6 and the probability that the third hits the target is 1/4 ?

Let A be the event that the first contestant hits the target and B be the second contestant hits the target and C be the contestant hits the target, then:
P(A)=1/3   →P(A^c )=2/3
P(B)=1/6   →P(B^c )=5/6
P(C)=1/4   →P(C^c )=3/4
Only one contestant hits the target mean that
 A∩B^c∩C^c    or  A^c∩B∩C^c   or  A^c∩B^c∩C
     Let D be the event that only one contestant the target, then:
       D=(A∩B^c∩C^c )∪(A^c∩B∩C^c )∪(A^c∩B^c∩C)
∴P(D)=P(A∩B^c∩C^c )+P(A^c∩B∩C^c )+P(A^c∩B^c∩C)
=P(A)P(B^c )P(C^c )+P(A^c )+P(B)+P(C^c )+P(A^c )P(B^c )P(C)
=(1/3)(5/6)(3/4)+(2/3)(1/6)(3/4)+(2/3)(5/6)(1/4)
=31/72
    P(A│D)=(P(A∩D))/(P(D))=(5⁄24)/(31⁄72)=15/31