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N " 5 2 1:1?) J ‘55” fqml I Solutions to the. midterm ~ ‘1
Pm gm (1 [771—9 STAT 120A, Spring 2003
51%) :1 '33“ Dept. of Statistics ' ’} a n3! .“3‘. _ ”VDJH’lO. 7T “51'! 'h snnro ”Hf" __ UCSB
May 8, 2008 "\ DO] 1. I will get a speciﬁc card if, after shufﬂing the Cards, it is number 1 or 5 ,n. ”P aw 5049 [3 or 9 or 49, i.e. if its number is 4k+1 (0 S k g 12). Hence, that the spade (b) ace comes to. me has 13/52=1f4 probability. Given that this
happens, the probability of getting the diamond (0) age is 12/51. And
simﬂerly, if those twa aces are with me, the heart ('59) ace. I get with 7 f
probability 11/50, and then the club (i) ate with probability 10/49. 30 the probability of getting all of them is The expected Win is therefore 100  0.0026+0_ ~ (1 — 0.0026) = 0.26, that 3 l
is, 26 cents. . 0] 2. (This is‘ thesaJne as the “hat problem” in the book.) The complement
is that at least one person sits in 1313/ her chair. If 14,3, 0,1) are the
7P respective events that person a, b, c, d site in his/her chair, then P(AUBUCUD)= 6’ {( P(A)+P(‘B)+P(C)+P(D)—P(AB)~P(AC’)—P(AD)—P(BC)WP(BD)—P(C’D)
 P +P(ABC‘) + P(ABD) + P(ACD) + PCBCD) — P(ABCD).
Since P(ABC') : P(ABC'D), the sought probability is 1 # 4P(A) + 6P(_AB) — 4P(.ABO) + P(ABCD) = 15p 1 — 4P(A) + 6P(AB) — 3P§ABOD) = . 2 1
1 — 4/4+ 6i.1 — 3i _ 9/24 _ 0.375. ...
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 Spring '09
 ENGLANDER
 Statistics, Probability

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