Chapter 10 Counting Methods and Probability - 10.6 Construct and Interpret Binomial Distributions - 10.6 Exercises - Skill Practice - Page 728: 33

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The standard formula: $P(number-of-successes)=\frac{n!}{(n-k)!k!}.p^k(1-p)^{n-k}$ Substituting $p=0.3,n=3,k=0$ we have: $P(0)=\frac{3!}{(3-0)!0!}.(0.3)^0(1-0.3)^{3-0}\approx0.343$ Substituting $p=0.3,n=3,k=1$ we have: $P(0)=\frac{3!}{(3-1)!1!}.(0.3)^1(1-0.3)^{3-1}\approx0.441$ Substituting $p=0.3,n=3,k=2$ we have: $P(0)=\frac{3!}{(3-2)!2!}.(0.3)^2(1-0.3)^{3-2}\approx0.189$ Substituting $p=0.3,n=3,k=3$ we have: $P(0)=\frac{3!}{(3-3)!3!}.(0.3)^3(1-0.3)^{3-3}\approx0.027$