Question

(1) For the random experiment “toss a fair coin twice and then roll a balanced die twice”, what’s the total count of the sample space? Let A be the event {both tosses show HEADs and both rolls show sixes}, B = {both tosses show the same side of the coin and both rolls show the same number}, C = {both tosses show the same side OR both rolls show the same number} and D = {the 1st roll of the die shows an odd number and the 2nd roll shows an even number}. Find P(A), P(B) P(C) and P(D). What are P(A|B) and P(A|C)?

Answer 1

The probability here is computed as:

P(A) = P(both tosses show heads and both rolls show sixes) = 0.5*0.5*(1/6)*(1/6) = 1/144

Therefore 1/144 = 0.006944 is the required probability here.

P(B) = P(both show the same side of the coin and both rolls the same numbers)

= P(both show heads and both show the same numbers) * P(both show tails and both show the same numbers)

= 0.5*0.5*(6/36) + 0.5*0.5*(6/36)

= 0.5*(1/6)

= 1/12

Therefore 1/12 = 0.0833 is the required probability here.

P(C) = P(both tosses show the same side or both rolls the same numbers )

= P(both tosses show same side ) + P(both dice throw shows the same number ) - P(both show the same side of the coin and both rolls the same numbers)

= 2*0.5*0.5 + (1/6) - (1/12)

= 0.5 + (1/12)

= 7/12

Therefore 7/12 = 0.5833 is the required probability here.

P(D) = P(1st roll of the die shows an odd number and second roll shows an even numbers )

= P(odd numbers on dice ) P(even number on die)

= 0.5*0.5

= 0.25

Therefore 0.25 is the required probability here.

The conditional probabilities are now computed here as:

P(A | B) = P(both show heads and both rolls show sixes | both both tosses show the same side of the coin and both rolls show the same number) is computed using Bayes theorem here as:

P(A | B) = P(A and B) / P(B) = P(A) / P(B) = (1/144) / (1/12) = 1/12

Therefore 1/12 = 0.0833 is the required probability here.

P(A | C) = P(both show heads and both rolls show sixes | both tosses show the same side OR both rolls show the same number ) is computed using Bayes theorem here as:

P(A | C) = P(A and C) / P(C) = P(A) / P(C) = (1/144) / (7/12) = 1/84

Therefore 1/84 = 0.0119 is the required probability here.