Question

An urn contains 6 white and 10 black balls. The figure gives by
the roll of a dice balance indicates the number of balls that will be drawn without delivery of the ballot box.
Let A be the event defined by:
A: all the balls drawn from the urn are white.
What is the probability that the dice has delivered a 3 knowing that A has realized (Use Bayes' Law)?

Answer 1

An urn contains 6 white and 10 black balls. The number given by the roll of a dice indicates the number of balls that will be drawn out of the urn without delivery back to the ballot box.

Let A be the event defined by:
A: All the balls drawn from the urn are white.

a)

What is the probability that the dice has delivered a 3 knowing that A has realized?

This question can be solved using Baye's Law which states that

If E₁, E₂, .............., E_n are n-mutually exclusive events with P(E_i) ≠ 0, for i = 1, 2, 3,.................., n. For any arbitrary event A which is a subset of such that P(A) > 0, then

Now, Here, A dice is rolled and the possible outcomes are 1,2,3,4,5,6. and all have probability =1/6

Let E_i denote the event that i comes on the die. And if number i comes on the die that many number of balls are drawn from the urn without replacement.

There are 6 white and 10 black balls in the urn. So, in total there are 16 balls.

Probability that a dice has delivered a 3 = P(E₃)= 1/6

P(A|E3) means,probability that all balls drawn are white(A), given 3 is obtained on dice which implies 3 balls are to be drawn from urn and all 3 of them should be white = P(A|E3) =

in numerator because all 3 balls are to be white, and in denominator because 3 balls are drawn from a urn containing 16 balls, both black and white.

Similarly it can be understood for rest outcomes 1,2,4,5,6

P(A|E1) =

P(A|E2) =

P(A|E4)=

P(A|E5)=

P(A|E3)=

Required probability using Baye's law,

= 0.065476