Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 2 / 17 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

i

Solution

The law of total probability weights each jar's red fraction by its selection probability:

P (R) = \frac{1}{2} \frac{20}{50} + \frac{1}{2} \frac{180}{350} = \frac{1}{5} + \frac{9}{35} = \frac{16}{35} .

The large difference in jar sizes does not change their equal selection probabilities; pooling their contents would describe a different experiment.

ii

Solution

By Bayes' theorem,

P (X ∣ R) = \frac{P ( R ∣ X ) P ( X )}{P ( R )} = \frac{( 20/50 ) ( 1/2 )}{16/35} = \frac{7}{16} .

The denominator sums red outcomes from both possible jar choices, as calculated in (i).

 Ancestors (12)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version