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--- b:head_first_statistics:geometric_binomial_and_poisson_distributions [2025/10/06 20:59] – [From a scratch (Proof of Binomial Expected Value)] hkimscil
+++ b:head_first_statistics:geometric_binomial_and_poisson_distributions [2025/10/06 23:43] (current) – [e.g.,] hkimscil
@@ Line 877: / Line 877: @@
 c <- choose(n,r)
 ans1 <- c*(p^r)*(q^(n-r))
-ans1
+ans1    # or
+choose(n, r)*(p^r)*(q^(n-r))
+dbinom(r, n, p)
 </code>
 <code>
 > p <- .25
@@ Line 888: / Line 894: @@
 > ans <- c*(p^r)*(q^(n-r))
 > ans
+[1] 0.2636719
+>
+> choose(n, r)*(p^r)*(q^(n-r))
+[1] 0.2636719
+>
+> dbinom(r, n, p)
 [1] 0.2636719
 >
 >
 </code>
 Ans 2.
@@ Line 903: / Line 920: @@
 ans2 <- c*(p^r)*(q^(n-r))
 ans2
+choose(n, r)*(p^r)*(q^(n-r))
+dbinom(r, n, p)
 </code>
 <code>
@@ Line 914: / Line 936: @@
 > ans2
 [1] 0.08789062
+>
+> choose(n,r)*(p^r)*(q^(n-r))
+[1] 0.08789062
+>
+> dbinom(r, n, p)
+[1] 0.08789063
+>
 >
 </code>
-Ans 3.
+Ans 3. 중요
 <code>
 ans1 + ans2
+dbinom(2, 5, .25) + dbinom(3, 5, .25)
+dbinom(2:3, 5, .25)
+sum(dbinom(2:3, 5, .25))
+pbinom(3, 5, .25) - pbinom(1, 5, .25)
 </code>
-<code>> ans1 + ans2
+<code>
+> ans1 + ans2
 [1] 0.3515625
+> dbinom(2, 5, .25) + dbinom(3, 5, .25)
+[1] 0.3515625
+> dbinom(2:3, 5, .25)
+[1] 0.26367187 0.08789063
+> sum(dbinom(2:3, 5, .25))
+[1] 0.3515625
+> pbinom(3, 5, .25) - pbinom(1, 5, .25)
+[1] 0.3515625
+>
 </code>
@@ Line 979: / Line 1022: @@
 > </code>
+Q. 한 문제를 맞힐 확률은 1/4 이다. 총 여섯 문제가 있다고 할 때, 0에서 5 문제를 맞힐 확률은? dbinom을 이용해서 구하시오.
+<code>
+p <- 1/4
+q <- 1-p
+n <- 6
+pbinom(5, n, p)
+- dbinom(6, n, p)
+</code>
+<code>
+> p <- 1/4
+> q <- 1-p
+> n <- 6
+> pbinom(5, n, p)
+[1] 0.9997559
+> 1 - dbinom(6, n, p)
+[1] 0.9997559
+</code>
+중요 . . . .
+<code>
+# http://commres.net/wiki/mean_and_variance_of_binomial_distribution
+# ##################################################################
+#
+p <- 1/4
+q <- 1 - p
+n <- 5
+r <- 0
+all.dens <- dbinom(0:n, n, p)
+all.dens
+sum(all.dens)
+choose(5,0)*p^0*(q^(5-0))
+choose(5,1)*p^1*(q^(5-1))
+choose(5,2)*p^2*(q^(5-2))
+choose(5,3)*p^3*(q^(5-3))
+choose(5,4)*p^4*(q^(5-4))
+choose(5,5)*p^5*(q^(5-5))
+all.dens
+choose(5,0)*p^0*(q^(5-0)) +
+  choose(5,1)*p^1*(q^(5-1)) +
+  choose(5,2)*p^2*(q^(5-2)) +
+  choose(5,3)*p^3*(q^(5-3)) +
+  choose(5,4)*p^4*(q^(5-4)) +
+  choose(5,5)*p^5*(q^(5-5))
+sum(all.dens)
+#
+(p+q)^n
+# note that n = whatever, (p+q)^n = 1
+</code>
+<code>
+> # http://commres.net/wiki/mean_and_variance_of_binomial_distribution
+> # ##################################################################
+> #
+> p <- 1/4
+> q <- 1 - p
+> n <- 5
+> r <- 0
+> all.dens <- dbinom(0:n, n, p)
+> all.dens
+[1] 0.2373046875 0.3955078125 0.2636718750 0.0878906250
+[5] 0.0146484375 0.0009765625
+> sum(all.dens)
+[1] 1
+>
+> choose(5,0)*p^0*(q^(5-0))
+[1] 0.2373047
+> choose(5,1)*p^1*(q^(5-1))
+[1] 0.3955078
+> choose(5,2)*p^2*(q^(5-2))
+[1] 0.2636719
+> choose(5,3)*p^3*(q^(5-3))
+[1] 0.08789062
+> choose(5,4)*p^4*(q^(5-4))
+[1] 0.01464844
+> choose(5,5)*p^5*(q^(5-5))
+[1] 0.0009765625
+> all.dens
+[1] 0.2373046875 0.3955078125 0.2636718750 0.0878906250
+[5] 0.0146484375 0.0009765625
+>
+> choose(5,0)*p^0*(q^(5-0)) +
++   choose(5,1)*p^1*(q^(5-1)) +
++   choose(5,2)*p^2*(q^(5-2)) +
++   choose(5,3)*p^3*(q^(5-3)) +
++   choose(5,4)*p^4*(q^(5-4)) +
++   choose(5,5)*p^5*(q^(5-5))
+[1] 1
+> sum(all.dens)
+[1] 1
+> #
+> (p+q)^n
+[1] 1
+> # note that n = whatever, (p+q)^n = 1
+>
+</code>
 ===== Proof of Binomial Expected Value and Variance =====
 [[:Mean and Variance of Binomial Distribution|이항분포에서의 기댓값과 분산에 대한 수학적 증명]], Mathematical proof of Binomial Distribution Expected value and Variance