Response to Exercise 2 of Introduction to Probability

Let \(S_{200}\) be the number of heads that turn up in 200 tosses of a fair coin. Estimate:

\(E(x) = np = 200 * 0.5 = 100\)

\(\sigma^{2} = \sqrt{npq}=\sqrt{200*0.5*0.5}=\sqrt{50}=5\sqrt{2}\)

var <- 5*sqrt(2)
  1. \(P(S_{200} = 100)\)

Equivalent to:

\(1 - P(S_{200}\le99) - P(S_{200}\ge101)\)

\(P(S_{200}\le99)\)

\(P(S_{200}\le99) = P(S_n^{*}\le \frac{99+0.5-100}{5\sqrt{2}})=P(S_n^{*}\le \frac{-0.5}{5\sqrt{2}})\) \(P(S_{200}\le99) =NA(-\infty,-\frac{0.5}{5\sqrt{2}})\)

pA1 <-pnorm(-0.5/var)

\(P(S_{200}\ge101) = 1 - P(S_{200}<101)\)

\(P(S_{200}\ge101) = P(S_n^{*}< \frac{101+0.5-100}{5\sqrt{2}})=P(S_n^{*}\le \frac{0.5}{5\sqrt{2}})\)

\(P(S_{200}\ge101) =NA(-\infty,\frac{1}{10\sqrt{2}})\)

pA2 <-1 - pnorm(0.5/var)

PA <- 1 - pA1 - pA2

\[P(S_{200} = 100) = 0.056372\]

  1. \(P(S_{200}=90) = P(S_{200}\le90) - P(S_{200}\le89)\)

\(P(S_{200}\le90) = P(S_n^{*}< \frac{90+0.5-100}{5\sqrt{2}})=P(S_n^{*}\le \frac{-9.5}{5\sqrt{2}})\)

\(P(S_{200}\le90) = NA(-\infty,-\frac{9.5}{5\sqrt{2}})\)

\(P(S_{200}\le89) = P(S_n^{*}< \frac{89+0.5-100}{5\sqrt{2}})=P(S_n^{*}\le \frac{-10.5}{5\sqrt{2}})\)$

\(P(S_{200}\le89) = NA(-\infty,-\frac{10.5}{5\sqrt{2}})\)

pB1 <-pnorm(-9.5/var)
pB2 <-pnorm(-10.5/var)

PB <- pB1 - pB2

\[P(S_{200} = 90) = 0.0207726\]

  1. \(P(S_{200}=80) = P(S_{200}\le80) - P(S_{200}\le79)\)

\(P(S_{200}\le80) = P(S_n^{*}< \frac{80+0.5-100}{5\sqrt{2}})=P(S_n^{*}\le \frac{-19.5}{5\sqrt{2}})\)

\(P(S_{200}\le90) = NA(-\infty,-\frac{19.5}{5\sqrt{2}})\)

\(P(S_{200}\le79) = P(S_n^{*}< \frac{79+0.5-100}{5\sqrt{2}})=P(S_n^{*}\le \frac{-20.5}{5\sqrt{2}})\)$

\(P(S_{200}\le89) = NA(-\infty,-\frac{20.5}{5\sqrt{2}})\)

pC1 <-pnorm(-19.5/var)
pC2 <-pnorm(-20.5/var)

PC <- pC1 - pC2

\[P(S_{200} = 80) = 0.0010394\]