Astudent answers a multiple choice exam question that has 4 possible answers. suppose that the probability that the student knows the answer to the question is 0.80 and the probability that the student guesses (doesn’t know the answer) is 0.20. if the student knows the answer, the probability of choosing the correct answer is 1 (because the student knows the answer). if the student guesses, the probability of choosing the correct answer is 0.25. a) what is the probability that the student really knew the correct answer? (that is, what is the probability that the student knew the answer, given that the student got the answer correct). b) suppose there is a third option. the student knew the answer with probability of .7, the student guessed with probability .2, and the student(incorrectly) thought they knew the answer given that the student thought they knew it is 0. the conditional probability of a student getting the correct answer given that they guessed is still .25. using this information, calculate the probability that the student really knew the correct answer.

15% chance.

step-by-step explanation:

3/10 times, the 5 year old will get a tablet. 5/10 times, the 13 yer old will get a tablet. multiply both fractions to get 15/100 times, both would have a tablet.

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step-by-step explanation:

69.64%

step-by-step explanation:

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.