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--- b:head_first_statistics:binomial_distribution [2025/10/07 06:42] – [Expectation and Variance of] hkimscil
+++ b:head_first_statistics:binomial_distribution [2025/10/07 06:45] (current) – [Proof of Binomial Expected Value and Variance] hkimscil
@@ Line 1: / Line 1: @@
-======= Binomial Distributions =======
+======= Binomial Distribution =======
   - 1번의 시행에서 특정 사건 A가 발생할 확률을 p라고 하면
@@ Line 115: / Line 115: @@
 \end{eqnarray*}
+====== e.g., ======
+<WRAP box>
+In the latest round of Who Wants To Win A Swivel Chair, there are 5 questions. The probability of
+getting a successful outcome in a single trial is 0.25
+  - What’s the probability of getting exactly two questions right?
+  - What’s the probability of getting exactly three questions right?
+  - What’s the probability of getting two or three questions right?
+  - What’s the probability of getting no questions right?
+  - What are the expectation and variance?
+</WRAP>
+Ans 1.
+<code>
+p <- .25
+q <- 1-p
+r <- 2
+n <-5
+# combinations of 5,2
+c <- choose(n,r)
+ans1 <- c*(p^r)*(q^(n-r))
+ans1    # or
+choose(n, r)*(p^r)*(q^(n-r))
+dbinom(r, n, p)
+</code>
+<code>
+> p <- .25
+> q <- 1-p
+> r <- 2
+> n <-5
+> # combinations of 5,2
+> c <- choose(n,r)
+> 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.
+<code>
+p <- .25
+q <- 1-p
+r <- 3
+n <-5
+# combinations of 5,3
+c <- choose(n,r)
+ans2 <- c*(p^r)*(q^(n-r))
+ans2
+choose(n, r)*(p^r)*(q^(n-r))
+dbinom(r, n, p)
+</code>
+<code>
+> p <- .25
+> q <- 1-p
+> r <- 3
+> n <-5
+> # combinations of 5,3
+> c <- choose(n,r)
+> ans2 <- c*(p^r)*(q^(n-r))
+> 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. 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
+[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>
+Ans 4.
+<code>
+p <- .25
+q <- 1-p
+r <- 0
+n <-5
+# combinations of 5,3
+c <- choose(n,r)
+ans4 <- c*(p^r)*(q^(n-r))
+ans4
+</code>
+<code>> p <- .25
+> q <- 1-p
+> r <- 0
+> n <-5
+> # combinations of 5,3
+> c <- choose(n,r)
+> ans4 <- c*(p^r)*(q^(n-r))
+> ans4
+[1] 0.2373047
+> </code>
+Ans 5
+<code>
+p <- .25
+q <- 1-p
+n <- 5
+exp.x <- n*p
+exp.x
+</code>
+<code>> p <- .25
+> q <- 1-p
+> n <- 5
+> exp.x <- n*p
+> exp.x
+[1] 1.25</code>
+<code>
+p <- .25
+q <- 1-p
+n <- 5
+var.x <- n*p*q
+var.x
+</code>
+<code>> p <- .25
+> q <- 1-p
+> n <- 5
+> var.x <- n*p*q
+> var.x
+[1] 0.9375
+> </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 Expected Value and Variance in Binomial Distribution ======
+[[:Mean and Variance of Binomial Distribution|이항분포에서의 기댓값과 분산에 대한 수학적 증명]], Mathematical proof of Binomial Distribution Expected value and Variance