There is a 30 percent chance that A can fix her busted computer. If A cannot, then there is a 40 percent chance that her friend B can fix it. a) Find the probability it will be fixed by either A or B.
b) If it is fixed, what is the probability it will be fixed by B.

(a) 0.70

(b) 0.40

Step-by-step explanation:

Mutually-exclusive events are those events which cannot occur together.

Consider that events X and Y are mutually exclusive.

P (X and Y) = 0

Here the two events can be defined as follows:

A = A can fix the busted computer

B = B can fix the busted computer

The information provided are as follows:

P (A) = 0.30

P (B) = 0.40

If A cannot, then there is a chance that her friend B can fix it.

The above statement suggest that the events A and B are mutually exclusive, i.e. if A can fix the computer then B does not have to and if cannot then only B will fix it.

That is, P (A and B) = 0.

(a)

Compute the probability it will be fixed by either A or B as follows:

P (A or B) = P (A) + P (B) - P (A and B)

                = 0.30 + 0.40 - 0

                = 0.70

Thus, the probability it will be fixed by either A or B is 0.70.

(b)

Compute the probability that if it is fixed it will be fixed by B as follows:

P (not A and B) = P (B) - P (A and B)

                         = 0.40 - 0

                         = 0.40

Thus, the probability that if it is fixed it will be fixed by B is 0.40.

y= 2x+7

changed to y= 5x +7

as slope is increased from 2 to 5

therefore it would be less steeper

so, option d) is correct

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.