1. Patient Outcomes Each patient will be evaluated at up to 4 time points (after 2, 4, 8 and 12 cycles) of therapy. Compared to the patient’s baseline evaluation, the patient’s disease will be categorizes as R = CR/PR (response), F = progressive disease or death (failure), or S (stable disease = neither R nor F) A patient with outcome R or F at any stage is scored as having that overall outcome A patient with outcome S is re-evaluated after subsequent cycles of therapy

2. Evaluation of Response Each patient will be assigned one of the following categories 1.Complete response 2.Partial response 3.Stable disease 4.Progressive disease 5.Early death for malignant disease 6.Early death from toxicity 7.Early death because of other causes 8.Unknown (not assessable insufficient data) Patients in response categories 4-9 should be considered as failing

3. Possible Patient Outcomes Eval 1Eval 2Eval 3Eval 4Overall Outcome 1RR 2FF 3SRR 4SFF 5SSRR 6SSFF 7SSSRR 8SSSFF 9SSSSR

4. Ewings – Current Data Eval 1Eval 2Eval 3Eval 4Number of Patients 1R0 2F7 3SR1 4SF1 5SSR0 6SSF1 7SSSR0 8SSSF0 9SSSS0 Pending patients: 1) S at Eval 1, Pending Eval 2 2) Pending Eval 1

5. Stopping Boundary How close are we in the Ewings subtype? Let  denote the overall response rate Accrual will be terminated if Pr(  > 0.35 | Data ) < 0.05. That is, there is less than a 5% chance that we have reached the targeted 35% overall response rate Currently with 2 pending patients enrolled Pr(  > 0.35 | Data ) = 0.098

6. Possible Results For Pending Patients By 11/16/2006 both pending patients are scheduled to have scans. Hypothetical outcomes are in maroon. CasePend. Pat.Eval 1Eval 2 Pr(  > 0.35 | Data ) 11SF0.035 2F 21SR0.160 2F 31SF0.104 2R 41SR0.316 2R 51SS0.087 2S

7. Pending Patients By 11/16/2006 both pending patients are scheduled to have scans. CasePend. Pat.Eval 1Eval 2 Pr(  > 0.35 | Data ) 11SF0.035 2F 21SR0.160 2F 31SF0.104 2R 41SR0.316 2R 51SS0.087 2S

8. Calculating  Let  R,j and  S,j denote the probability of R and S at stage j respectively.  =  R,1 +  S,1  R,2 +  S,1  S,2  R,3 +  S,1  S,2  S,3  R,4 +  S,1  S,2  S,3  S,4