Math 230 Homework 1

# Math 230 Homework 1 - P(0 = 1/2 P(1 = 1/6 P(2 = 1/12 P(3 =...

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Math 230 Homework 1 Orhun Alp Oral 20801143 Section-3 1.) From hyper geometric distribution formula : C(n,k) x C(N-n,m-k) / C(N,m) 2.) P(A) = 0.5, P(B) = 0.5, P(C) = 1/6 P(A B) = 0.25 = P(A)xP(B)   independent P(A C) = 0.5 x 1/6 = P(A)xP(C)   independent P(B C) = 0.5 x 1/6 = P(B)xP(C)   independent P(A B C) ={1-6,2-5,3-4} = 1/6 x 1/6 x 3 = 1/12 ≠ P(A)xP(B)xP(C) = 1/24 not independent 3.) P(defective) = 0.02 , P(not-defective) = 0.98 P(TesterDefective|defective) = 0.94 , P(TesterNotDefective|not-defective) = 0.95 P(defective|TesterDefective) = ? = [ P(TesterDefective |defective) x P(defective) ] / P(TesterDefective) = [0.94 x 0.02 ] / [P(TesterDefective defective) + P(TesterDefective not- defective) ] = [0.94 x 0.02 ] / [0.94x0.02 + 0.05 x 0.98] = 0.0188 / [0.0188 + 0.049] = 0.2772 4.) a.) Let P(X) is the probility of the first player’s win for x times.

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Unformatted text preview: P(0) = 1/2, P(1) = 1/6, P(2) = 1/12, P(3) = 1/5, P(4) = 1/5 b.) E[X] = ∑x.P(x) = 0x1/2 + 1x1/6 + 2x1/12 + 3x1/5 + 4x1/5 = 1/3 + 3/5 + 4/5 = 26/15 5.) P(win) = P(sum is 7 or 11) = 8/36 P(lose) = P(sum is 2,3 or 12) = 4/36 P(continue) = 24/36 f(x) = {x=0: 4/36 ; x=1: 8/36 ; x=2 : 24/36 } F x (t) = {t<0: 0 ; 0<=t<1: 4/36 ; 1<=t<2 : 12/36 ; t>=2 : 1} 6.) a.) 0.7 x ( 1 + 0.3 + (0.3) 2 + (0.3) 3 ) = 0.7 x (1 + 0.3 + 0.09 + 0.027) = 0.974 b.) 0.7 x ( 1 + (0.3) 2 + (0.3) 4 ...) = 0.7 x 1/1-(0.3) 2 = 0.79 7.) a.) 0.99 10 = 0.904 b.) (0.99 10 ) 5 = 0.99 50 = 0.604 8.) M x (t) = E(e tx ) = ∑etx.C(n,x).p x .(1-p) n-x = ∑ C(n,x).(pe t ) x (1-p) n-x [M x (t=0)]’ = np = E(x) [M x (t=0)]’’ =np(1-p)...
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## This note was uploaded on 02/19/2011 for the course MATH 315 taught by Professor Selcuk during the Fall '10 term at Bilkent University.

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Math 230 Homework 1 - P(0 = 1/2 P(1 = 1/6 P(2 = 1/12 P(3 =...

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