chapter 1

Chapter 2: Probabilityd. For what values of p will P(G | R 1 .5,P(G | R 3 < R 1 < R 2 ) > .5?.6p.6p.6p+ .1(1 − p).1+.5pP(G | R 1 < R 2 < R 3 ) = = > . 5 iff.25p.25p + .2(1 − p).15p> ..15p + .7(1 − p)P(G | R 1 < R 3 < R 2 ) = > . 5 iff4p >914>171p >7P(G | R 3 < R 1 < R 2 ) = 5 iff p (most restrictive)If14>17p always classify as granite.105. P(detection by the end of the nth glimpse) = 1 – P(not detected in 1 st n)= 1 – P(G 1 ′ ∩ G 2 ′ ∩ … ∩ G n ′ ) = 1 - P(G 1 ′)P(G 2 ′) … P(G n ′)= 1 – (1 – p 1 )(1 – p 2 ) … (1 – p n ) = 1 - π (1 − p )ni=1i106.a. P(walks on 4 th pitch) = P(1 st 4 pitches are balls) = (.5) 4 = .0625b. P(walks on 6 th ) = P(2 of the 1 st 5 are strikes, #6 is a ball)= P(2 of the 1 st 5 are strikes)P(#6 is a ball)= [10(.5) 5 ](.5) = .15625c. P(Batter walks) = P(walks on 4 th ) + P(walks on 5 th ) + P(walks on 6 th )= .0625 + .15625 + .15625 = .375d. P(first batter scores while no one is out) = P(first 4 batters walk)=(.375) 4 = .0198107.a. P(all in correct room) = = = . 041714 × 3×2 × 1124b. The 9 outcomes which yield incorrect assignments are: 2143, 2341, 2413, 3142, 3412,9 =243421, 4123, 4321, and 4312, so P(all incorrect) = . 37593