Problem 1

P(R) = 0.2 P(N) = 0.3
P(R^' N^' ) = 0.6

Part (a)
P(R N) = 1 - P(R^' N^' )
                = 1 - 0.6 = 0.4

Part (b)
P(R N) = P(R) + P(N) - P(R N)
                = 0.2 + 0.3 - 0.4 = 0.1

Part (c)
Events R and N are not mutually exclusive since P(R N) 0.

Part (d)
P(R) · P(N) = (0.2) (0.3) = 0.06 P(R N)
Hence events R and N are not independent

Problem 2

A - customer purchases a suit
H - customer purchases a shirt
T - customer purchases a tie
P(A) = 0.22    P(H) = 0.30     P(T) = 0.28

P(A H) = 0.11    P(A T) = 0.14    P(H T) = 0.10
P(A H T) = 0.06

Part (a)
P(purchases at least one item) = P (A H T )
             = P(A) + P(H) + P(T) - P(A H) - P(A T) - P(H T) + P(A H T)
             = 0.22 + 0.30 + 0.28 -0.11 - 0.14 -0.10 + 0.06 = 0.51
P(purchases none of these items) = P(A^' H^' T^' ) = 1 - P (A H T )
                                                   = 1 - 0.51 = 0.49

Part (b)

Problem 3

E - Englishman             A - American           V - selected word is a vowel

Rigour                              Rigor

P(V | E) = 3/6 = 0.5       P(V | A) = 2/5 = 0.4
P(E) 0.4                         P(A) = 0.6
P(E | V) = P(E V) / P(V)
P(V) = P(V | E) P(E) + P(V | A) P(A)
         = (0.5)(0.4) + (0.4) ( 0.6)
         = 0.20 + 0.24 = 0.44
P(E | V) = 0.2/0.44 = 0.45
 

Problem 4

E_i = engine i operates
P(E_i ) = 0.9

P("safe flight") = P( at least one engine operates) = P(E₁ E₂ E₃ . . . . .E_n )
                      = 1 - P(none of the engines operates)
                      = 1 - P(E₁^' E₂^' E₃^' . . . . .E_n^')
                      = 1- P(E₁^' ) P( E₂^' ) P(E₃^' ) . . . . .P( E_n^' ) since independent engines
                       =1 - (0.1)ⁿ = 0.999
(0.1)ⁿ = 0.001
(1/10)ⁿ = (1/10)³
Hence n = 3
 

Problem 5

C_i - an individual contributed virus to the pool
Let P(C_i^' ) = q
P(batch contaminated) = P( at least one individual contributed virus)
                                   = 1 - P(none contributed virus)
                                    = 1 - P(C₁^' C₂^' C₃^' . . . . .C₁₀₀^')
                                   = 1- P(C₁^' ) P( C₂^' ) P(C₃^' ) . . . . .P( C₁₀₀^' ) assuming independence
                                   =1 - q¹⁰⁰
                                   =0.33
hence q = (0.33)^1/100 = 0.98897
P(an individual contributed virus) = 1- q = 1- 0.98897 = 0.01103
 
 

Problem 6

Five (5) tagged and fifteen (15) not tagged

Problem 7

R - right key W - wrong key

A - opens the door on the 4^th try

Part a : discarding previously used keys

Using counting rules

P(A) = (10 -1)!/10! = 1/10 = 0.1

Using conditional probability

P(W₁ W₂ W₃ R₄ )
             = P(W₁ ) P(W₂ | W₁ ) P(W₃ | W₁ W₂ ) P(R₄ | W₁ W₂ W₃ )
             = (9/10) (8/9) (7/8)(1/7) =1/10

Part b: not discarding previously used keys

Using counting rules

P(A) = (9³ )/ ( 10⁴ ) = 0.0729

Using independence rule

P(W₁ W₂ W₃ R₄ ) = P(W₁ ) P( W₂ ) P(W₃ ) P( R₄ )
                  = (9/10) (9/10) (9/10) (1/10) = 0.0729
 

Problem 8