For the dust bears,  among the F₂ progeny of  the initial cross of aaBBCCDD x AAbbCCDD there should be four phenotypes in a 9:3:3:1 ratio.  Since each dust bear produces 64 males and 64 females we expect the following numbers of dust bears of each type.  ( The expected number of progeny of each sex is determined by multiplying the expected frequency by the total number of progeny)
 

However, it is possible that there are greater or less than 4 males of the desired genotype produced from this cross.  Since it is essential to obtain at least one male dust bear with the phenotype: Blue Eye, solid, green pellicle and two terminal claws,  it is important to know the probability of obtaining zero males of the correct genotype.  This can be done with the binomial equation.

We can assign the 5 variables that go into the binomial expansion based on information given in the problem and our knowledge of Mendelian genetics.

We could also estimate the probability of obtaining 1, 2 , 3, etc. males in the F₂ cross using the binomial expansion. For one male, w = 1 and x - 63, for two males w = 2 and x = 63, etc).
 

 
The chance is very good that one or more males will be obtained in a single F₂ cross.   However, many geneticists would rather spend money than time so they might set up 2 or more F₂ crosses rather than risk a 1.6% failure rate.  For to crosses the change of getting zero males in the first cross and zero males in the second cross is .016 x .016 or about .026%.  With two or three crosses the risk of getting no males becomes quite small.
 

The number of F₂ crosses (Cross 4) needed to obtain at least one male with the final target phenotype of Blue Eye, solid, green pellicle and four terminal claws is more important.  The critical values for the binomial expansion are:

 

n	the total number of events is 64.  This is given in the problem.
w	is the total number of dust bears with the phenotype Blue Eye, solid, green pellicle and four terminal claws.  The number of these dust bears is  zero (0) (because we wish to know how unlikely this is).
x	is the total number of dust bears that do not have  the phenotype Blue Eye, solid, green pellicle and four terminal claws.  Since there are 64 total and none of them have the desired phenotype this value is 64.  (remember n = w + w , therefore 64 = 0 + 64)
p	The probability of obtaining a dust bear with the phenotype: Blue Eye, solid, green pellicle and four terminal claws is 1/64
q	The probability of obtaining a dust bear that does not do not have  the phenotype Blue Eye, solid, green pellicle and four terminal claws is  63/64.

Practically speaking, the number of crosses to be carried out is a function of the cost in terms of time and money if one fails to obtain the animals with the correct genotype.  Dust bears are inexpensive to rear, have large number of progeny and have a short generation time.  A mistake would be trivial.  Mice on the other hand have generation time of 3-4 months, are expensive to rear and have litters of 4 to 10.  Mistakes with mice could delay an experiment for more that a year. 

It is not enough to get a males of the desired genotype if you don't also get a females of the desired phenotype. Since both sex and genotype are independent events, the probability of not getting both a male and a female is given below.

In this dust bear example,  one would have to exceptionally lucky to get both a male and a female of the desired phenotype from a single F₂ cross (Cross 4).  If one set up 5 F₂ crosses, there is still greater than a 1% chance that one fails to obtain a male or a female or both.  With 10 crosses the probability that one fails to obtain a male or a female or both is .008%.