<prompt> Chance of passing a genetic disease y trait by the affected parents to children is 0.16. They plan to have two children. Probability of both the children having y trait is ?zero0.160.320.0256  <response>: Ans. is d i.e., 0.0256 As the chance of passing a genetic disease y trait by the affected parent is 0.16, every child has 0.16 16 probability of getting y trait. Because parents are planning to have two children: i Probability of 1st child having y trait P A 0.16 16 ii Probability of 2nd child having y trait P B 0.16 16 To know the probability of both children having y trait, multiplication law is applicable as both events are independent to each othernot mutually exclusive occurence of y trait in one child does not exclude the probability of having y trait in other P1 P A x P B 0.16 x 0.16 0.0256 2.56 Following facts are also impoant related to this questions: Probability of 1st child not having y trait, P C 0.84 or 84 1 - 0.16 a Probability of 2nd child not having y trait, P D 0.84 Therefore, other three possibilities beside the one asked in question: i Probability of both children not having y trait: P2 P C x P D 0.84 x 0.84 0.7056 70.56 ii Probability of 1st child having y trait and 2nd child not having y trait P3 P A x P D 0.16 x 0.84 0.1344 13.44 iii Probability, of 1st child not having y trait and 2nd child having y trait: 134 P C x P B 0.84 x 0.16 0.1344 13.44 Total Probability Pibta P1 P2 P3 P4 0.0256 0.7056 0.1344 0.1344 1 100 