A new surgical procedure is said to be successful 60% of the time. Suppose the operation is performed nine times and the results are assumed to be independent of one another. What are the probabilities of these events? (Round your answers to three decimal places.)

Answer:The probabilities are:For no successful surgeries: practically 0For one successful surgery: 0.004For two successful surgeries: 0.021For three successful surgeries: 0.074For four successful surgeries: 0.167For five successful surgeries: 0.251For six successful surgeries: 0.251For seven successful surgeries: 0.161For eight successful surgeries: 0.06For nine successful surgeries: 0.01Step-by-step explanation:Lets call X the total number of success. X counts the number of success from the same experiment repeated 9 times with a probability of success of 0.6 and one experiment independent of the other. Therefore X has Binomial distribution, X ≈ Bi(9,0.6).The range of X is {0,1,2,3,4,5,6,7,8,9} and the probability of X being equal to a value k in its range is the number [tex] P_X(k) [/tex]  given by [tex]P_X(k) = {9 \choose k} \, 0.6^k * (1-0.6)^{9-k}[/tex]Thus,[tex]P_X(0) = {9 \choose 0} (0.4)^9 = (0.4)^9 = 0.000262 , [/tex] rounded to 0[tex]P_X(1) = {9 \choose 1} 0.6 * 0.4^8 = 9 * 0.6 * 0.4^8 = 0.004[/tex][tex]P_X(2) = {9 \choose 2} 0.6^2 * 0.4^7 = 36 * 0.6^2*0.4^7 = 0.021 [/tex][tex]P_X(3) = {9 \choose 3} 0.6^3 * 0.4^6 = 84 * 0.6^3*0.4^7 = 0.074 [/tex][tex]P_X(4) = {9 \choose 4} 0.6^4 * 0.4^5 = 126*0.6^4*0.4^5 = 0.167 [/tex][tex]P_X(5) = {9 \choose 5} 0.6^5 * 0.4^4 = 126*0.6^5*0.4^4 = 0.251 [/tex][tex]P_X(6) = {9 \choose 6} 0.6^6 * 0.4^3 = 84*0.6^6*0.4^3 = 0.251 [/tex][tex]P_X(7) = {9 \choose 7} 0.6^7 * 0.4^2 = 36*0.6^7*0.4^2 = 0.161 [/tex][tex]P_X(8) = {9 \choose 8} 0.6^8*0.4 = 9*0.6^8*0.4 = 0.06 [/tex][tex]P_X(9) = {9 \choose 9} 0.6^9 = 0.6^9 = 0.01 [/tex]   I hope that works for you!