Probability Solutions Manual | Ross 8th Ed Sample Chapter, Exams of Science education

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1. Let R and N denote the events, respectively, that the student wears a ring and wears a necklace.

(a) P ( R ∪ N ) = 1 − .6 =.

(b) .4 = P ( R ∪ N ) = P ( R ) + P ( N ) − P ( RN ) = .2 + .3 − P ( RN ) Thus, P ( RN ) =.

1. Let A be the event that a randomly chosen person is a cigarette smoker and let B be the event that she or he is a cigar smoker.

(a) 1 − P ( A ∪ B ) = 1 − (.07 + .28 − .05) = .7. Hence, 70 percent smoke neither.

(b) P ( AcB ) = P ( B ) − P ( AB ) = .07 − .05 = .02. Hence, 2 percent smoke cigars but not cigarettes.

1. (a) P ( S ∪ F ∪ G ) = (28 + 26 + 16 − 12 − 4 − 6 + 2)/100 = 1/ The desired probability is 1 − 1/2 = 1/2.

(b) Use the Venn diagram below to obtain the answer 32/100.

S F

G

(c) since 50 students are not taking any of the courses, the probability that neither one is taking a course is

⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ = 49/198 and so the probability that at least one is taking a course is 149/198.

1. (a) 20, (b) 12, (c) 11, (d) 68, (e) 10,

I II

III

14. P ( M ) + P ( W ) + P ( G ) − P ( MW ) − P ( MG ) − P ( WG ) + P ( MWG ) = .312 + .470 + .525 − .086 −

1. (a)

(b)

(c)

(d)

(e)

1. (a) 6 5 4 3 2 5 6

⋅ ⋅ ⋅ ⋅ (^) (b) 5

⎜ ⎟ ⋅^ ⋅

⎝ ⎠ (^) (c) 5

(d)

⋅ ⋅ ⎛^ ⎞

(e) (^5)

⋅ ⎛^ ⎞

(f) (^5)

⋅ ⎛^ ⎞

(g) (^65) 6

(^8 ) 1 64 63 58

i = i ⋅ ⋅⋅⋅

1. Let A be the event that you are dealt blackjack and let B be the event that the dealer is dealt blackjack. Then,

P ( A ∪ B ) = P ( A ) + P ( B ) − P ( AB )

where the preceding used that P ( A ) = P ( B ) = 2 × 4 16 52 51

. Hence, the probability that neither is dealt blackjack is .9017.

1. (a) (^ 1)^ (^ 1) ( )( 1)

n n m m n m n m

(b) Putting all terms over the common denominator ( n + m )^2 ( n + m − 1) shows that we must prove that n^2 ( n + m − 1) + m^2 ( n + m − 1) ≥ n ( n − 1)( n + m ) + m ( m − 1)( n + m )

which is immediate upon multiplying through and simplifying.

1. (a)

(b)

(c)

1. P ({complete} = P {same} =

32. (^ 1)!

g b g g b g b g

1. (a)

(b)

(c)

⎜ ⎟ +^ ⎜ ⎟ +⎜ ⎟

(d) (^3 3 3 3 3 )

P R B P R P B P R B

1. (a)

⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ ≈^ .0045,

(b)

⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ = 1/17^ ≈^.

1. (a)

⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ = 1/12^ ≈^.

(b)

⎛ ⎞ ⎛ n ⎞ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ or^ n ( n^ −^ 1) = 12 or^ n^ = 4.

39. 5 4 3^12

1. ( 1) n^ m^ / n

n n N m

1. (a) 20 18 16 14 12 10 8 6 20 19 18 17 16 15 14 13

(b)

1. Let Ai be the event that couple i sit next to each other. Then

4 2 3 4 1

( ) 4 2 7!^6 2 6!^42 5!^2 4!

i i 8! 8! 8! 8!

P ∪ = A = ⋅^ − ⋅^ + ⋅^ − ⋅

and the desired probability is 1 minus the preceding.

1. P ( S ∪ H ∪ D ∪ C ) = P ( S ) + P ( H ) + P ( D ) + P (C) − P ( SH ) − … − P ( SHDC )

⎜ ⎟ −^ ⎜ ⎟+

1. (a) P ( S ∪ H ∪ D ∪ C ) = P ( S ) + … − P ( SHDC )

⎜ ⎟ −^ ⎜ ⎟ +^ ⎜ ⎟ −⎜ ⎟

(b) P (1 ∪ 2 ∪ … ∪ 13) =

1. Player B. If Player A chooses spinner (a) then B can choose spinner (c). If A chooses (b) then B chooses (a). If A chooses (c) then B chooses (b). In each case B wins probability 5/9.

M N

k r k M N r

1. P ( E 1 … En ) ≥ P ( E 1 … En − 1 ) + P ( En ) − 1 by Bonferonni’s Ineq.

1 1

n P Ei

−

∑ −^ ( n^ −^ 2) +^ P ( En )^ −^ 1 by induction hypothesis

n m n r r k r n m n m k k

⎜ ⎟ +^ −^ +

1. Let y 1 , y 2 , …, yk denote the successive runs of losses and x 1 , …, xk the successive runs of wins. There will be 2 k runs if the outcome is either of the form y 1 , x 1 , …, yk xk or x 1 y 1 , … xk , yk where all xi , yi are positive, with x 1 + … + xk = n , y 1 + … + yk = m. By Proposition 6.1 there are 1 1 2 1 1

n m k k

⎜⎝ (^) − ⎟ ⎜⎠ ⎝ (^) − ⎟⎠ number of outcomes and so

P {2 k runs} =

n m m n k k n

⎛ −^ ⎞ ⎛ −^ ⎞ ⎛ + ⎞

There will be 2 k + 1 runs if the outcome is either of the form x 1 , y 1 , …, xk , yk , xk +1 or y 1 , x 1 , …,

yk , xk yk + 1 where all are positive and ∑ xi = n , ∑ yi = m. By Proposition 6.1 there are

n m k k

⎜⎝ ⎟ ⎜⎠ ⎝ (^) − ⎟⎠ outcomes of the first type and^

n m k k

⎜⎝ (^) − ⎟ ⎜⎠ ⎝ ⎟⎠ of the second.