Of the members of the Spring Lake Bowling Lanes, 57% have a lifetime membership
and bowl regularly (three or more times a week). If 70% of the club members bowl
regularly, find the probability that a randomly selected member is a lifetime member,
given that he or she bowls regularly.
a) 0.056
b) 0.814
c) 0.533
d) 0.538

The probability that a randomly selected member is a lifetime member,  given that he or she bowls regularly is 0.814, option b

Step-by-step explanation:

Let "M" denotes the event that a club member has a lifetime membership and "B" be denotes the event that the club member bowls.

57% of the members have lifetime membership and bowl regularly. This means, the probability that a randomly selected member has a lifetime membership and bowls regularly =

70% of the club members bowl. This means, the probability that a randomly chosen member bowls = P(B) = 70% = 0.70

We have to find the probability that a randomly selected member is a lifetime member,  given that he or she bowls regularly. This is conditional probability which can be expressed mathematically as P(M | B)

The formula to calculate the conditional probability is:

So, the formula for  our case will be:

Using the values in this formula we get:

This means, the probability that a randomly selected member is a lifetime member,  given that he or she bowls regularly is 0.814

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.

a

step-by-step explanation:

the denominator cannot be the same as one of the factors.

8 denominator times 2 denominator = 16

and 7 numerator time 11 = 77

77/16= 77/16 ( they cannot be further simplified)