Chance of passing a genetic disease &;y&; trait by the affected parents to children is 0.16. They plan to have two children. Prabability of both the children having &;y&; trait is -
<p> In the given question, chance of passing a Genetic disease &;y&; trait by the affected parents to children is 0.16 and they plan to have two children, Probability of 1st child having &;y&; trait P(A)=0.16(16%). Probability of 2nd child having &;y&; trait,P(B)= 0.16(16%). Probability of both the children having &;y&; trait is. P( total)=P(A)+P(B). P(total)=0.16*0.16=0.0256(2.56%). Reference:Simple Biostatistics by Indrayan & Indrayan,1st edition,pg no:105-107. <\p>
zero
0.16
0.32
0.0256
Biostatistics