Question

1.a We have 12 dice: 8 are regular and 4 are irregular. The probability of getting a 3 with an irregular dice is twice the probability of anyone of the rest of the numbers. 1) Find the probability of getting a 3 2) If we have got a 3, find the probability of being tossed with a regular dice 3) Find the probability of getting a 3 with an irregular dice (0.8 points)

1.b Which one is true and why? P(A|B) + P(B|A) = 1 or P(A|B) + P(Ac|B) = 1 (0.2 points)

Answer 1

We are given here that:
P(3 | irregular) = 2P( x | irregular) for any x = 1, 2, 4, 5, or 6

Therefore 5P(x | irregular) + P(3 | irregular) = 1

P(x | irregular) = 1/7 , for x = 1, 2, 4, 5, 6
P( 3 | irregular) = 2/7

Also, we have 8 regular and 4 irregular die, therefore:
P( irregular) = 1/3 and P(regular) = 2/3

1) Using law of total probability, we get here:
P( 3) = P(3 | regular)P(regular) + P( 3 | irregular)P(irregular) = (1/6)*(2/3) + (2/7)*(1/3) = (1/9) + (2/21) = (7 + 6)/63 = 13/63

Therefore 13/63 is the required probability here.

2) Given that a 3 was tossed, probability that it was tossed through a regular die is computed using Bayes theorem here as:

P( regular | 3) = P(3 | regular)P(regular) / P(3) = (1/6)*(2/3) / (13/63) = 7/13

Therefore 7/13 is the required probability here.

3) Probability of getting a 3 with an irregular die is computed here as;

= P(irregular)P(3 | irregular) = (1/3)*(2/7) = 2/21

Therefore 2/21 is the required probability here.

1b) P(A | B) + P(B | A) = P(A and B) / P(B) + P(B and A) / P(A)

This may or may not be equal to 1 depending on the situation. Therefore this statement is False here.

P(A | B) + P(Ac | B) , given that B has occured, A can either occur or not occur. Therefore the sum of probability here should be 1. Therefore the given statement is True here.