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 ?
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 other/not 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%)
zero
0.16
0.32
0.0256