1. The probability for each child being a male or a female is ½. For five children, you would multiple by ½ five times, giving you 1/32. Because we are not assuming the order of the children, for example the first child being a female and the other four being males, we need to multiple 1/32 by 5.
This gives us our final answer of 5/32.2. The probability of two carriers having a normal child is ¾. Let’s say each parent has a genotype of Aa. A normal child will either have the genotype AA or Aa. There are three possibilities to have a normal child: 1) both parents give A; 2) mom gives A and dad gives a; or 3) mom gives a and dad gives A. For option 1, each parent has ½ of a chance of giving A. ½ * ½ = ¼.For option 2, mom has a ½ chance of giving A and dad has a ½ chance of giving a. ½ * ½ = ¼.For option 3, mom has ½ a chance of giving a and dad has ½ a chance of giving A. ½ * ½ = ¼.Since these events are all mutually exclusive, we add the final probabilities together: ¼ + ¼ + ¼ = ¾ chance of the child being normal.For a cystic child, the genotype must be aa. The only way for this to happen is if both parents give aa. Each parent has a ½ chance of giving an a. ½ * ½ = ¼.Again, as in question 1, we are not assuming the order of which child is born first – the normal one or the affected one. So, we multiple ¾ and ¼ by 2.2* ¼ * ¾ = 6/16, which is our final answer.