Consider the following cross: parent 1: yytt parent 2: yytt using the rules of probability, determine the probability that the offspring will show yytt genotype?

The answer is 1/8. Let's separate two traits. We have two events occurring together so using multiplication rule we will multiply probabilities of those events. Event 1: the probability that the offspring will show Yy genotype. In the cross of parents Yy and Yy, one of four offspring will have YY genotype, two of four offspring will have Yy genotype, and one of four offspring will have yy genotype. Thus, the probability that the offspring will show Yy genotype is 2/4. Event 2: the probability that the offspring will show TT genotype. In the cross of parents Tt and Tt, one of four offspring will have TT genotype, two of four offspring will have Tt genotype, and one of four offspring will have tt genotype. Thus, the probability that the offspring will show TT genotype is 1/4. Hence, the probability that the offspring will show YyTT genotype is: 2/4 * 1/4 = 2/16 = 1/8.