Consider the trial on which a 3 is first observed in successive rolls of a six-sided die. let a be the event that 3 is observed on the first trial. let b be the event that at 3.assumingthat each side has probability 1/6,

1. p(a)
2. p(b)
3.p(a u b)

1) 0.1667

2) 0.8333

3) 1

Step-by-step explanation:

Let A be the event that 3 is observed on the first trial.

Let B be the event that at least two trials are required to observe a 3

a) A six-sided die has 6 possible outcomes of 1,2,3,4,5,6

3 is one of the six possible outcomes

P(A) = no of favourable outcomes /no of possible outcomes

P(A) = 1/6 = 0.1667

b) The event of requiring at least two trials to observe a 3 is the event of not observing a 3 on the first trial.

P(B) = P(A')

P(A')= 1 - P(A)

P(B) = 1 - 1/6

= 5/6 = 0.8333

c) P(AnB) = P(A) + P(B)

= 1/6 + 5/6

= 6/6 = 1

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2

555.99.

step-by-step explanation:

hope this