Chapter 10 Counting Methods and Probability - 10.6 Construct and Interpret Binomial Distributions - 10.6 Exercises - Problem Solving - Page 729: 46a

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The standard formula: $P(number-of-successes)=\frac{n!}{(n-k)!k!}.p^k(1-p)^{n-k}$ Substituting $p=0.3,n=10,k=0$ we have: $P(0)=\frac{10!}{(10-0)!0!}.(0.3)^0(1-0.3)^{13-0}\approx0.0283$ Substituting $p=0.3,n=10,k=1$ we have: $P(1)=0.121$ Substituting $p=0.3,n=10,k=2$ we have: $P(2)=0.2335$ Substituting $p=0.3,n=10,k=3$ we have: $P(1)=0.2668$ Substituting $p=0.3,n=10,k=4$ we have: $P(4)=0.2001$ Substituting $p=0.3,n=10,k=5$ we have: $P(1)=0.1029$ Substituting $p=0.3,n=10,k=6$ we have: $P(6)=0.0368$ Substituting $p=0.3,n=10,k=7$ we have: $P(1)=0.009$ Substituting $p=0.3,n=10,k=8$ we have: $P(1)=0.00145$ Substituting $p=0.3,n=10,k=9$ we have: $P(1)=0.0001378$ Substituting $p=0.3,n=10,k=10$ we have: $P(10)=0.0000059$