The probability that any cutting takes root is 0.60. Either it takes root, or it doesn't--a binomial distribution. So for each cutting, p = 0.6 and q = p^-1 = 1 - p = 0.4. So for each pot, let n = 3. Iterating over the cuttings in each pot using the BInomial Distribution Formula:
P(n, k, p) = C(n,k) p^k * (1 - p)^(n-k),
where n is umber of trials, k is number of desired outcomes and p is the probability of that outcome in independent trials. So...
The probability of k = 0 successes in n = 3 trials while probability of success is p = 0.6 is
C(3,0) * ( (0.6) ^ 0) * ((0.4)^(3 - 0)) = 0.064
The probability of k = 1 successes in n = 3 trials while probability of success is p = 0.6 is
C(3,1) * ( (0.6)^1) * ((0.4)^(3 - 1)) = 0.288
The probability of k = 2 successes in n = 3 trials while probability of success is p = 0.6 is
C(3,2) * ( (0.6)^2) * ((0.4)^(3 - 2)) = 0.432
The probability of k = 3 successes in n = 3 trials while probability of success is p = 0.6 is
C(3,3) * ( (0.6)^3) * ((0.4)^(3 - 3)) = 0.216
(Notice also that the you can check your work by verifying that the TOTAL PROBABILITY for this distribution over [0:n] is
(k=0,k=n) ∑ C(n,k) p^k * (p-1)^(n-k)
= 0.064 + 0.288 + 0.432 + 0.216
= 1.)
So for 100 pots, we would expect 100*0.064 = 6.4 --> 6 pots to have 0 rooted cuttings,
100 * 0.288 = 28.8 -- > 29 pots to have exactly 1 rooted cutting,
100 * 0.432 = 43.2 --> 43 pots to have exactly 2 rooted cuttings and
100 * 0.216 = 21.6 --> 22 pots to have exactly 3 rooted cuttings.