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# GATE | GATE CS 2012 | Question 31

Suppose a fair six-sided die is rolled once. If the value on the die is 1, 2, or 3, the die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?
(A) 10/21
(B) 5/12
(C) 2/3
(D) 1/6

Explanation:

Solution set: { 6, (1,5), (1,6) ……}

i.e. P(6 appeared on first throw) +

P(1 appeared on first throw and 5 appeared on second throw) +

P(1 appeared on first throw and 6 appeared on second throw) + ………………..

= 1/6 + (1/6)(1/6) + (1/6)(1/6) + …..

= 1/6 + 9/36

= 5/12.

Viewpoint 2:

P(……) =   P(6 came on first throw) +  P(sum>= 6 and 1,2,3 appeared in first throw)

=      1/6                                +                 ????

P(1,2,3 appeared in first throw) = 1/2                                                   //P(E1)

P(sum >= 6 | 1,2,3 appeared in first throw) = 9/18                             //P(E2 | E1)

// Our new sample space is:  { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6) }

// 9 favorable cases: {(1,5), (1,6), (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6) }

P(sum>= 6 and 1,2,3 appeared in first throw) = (1/2)(9/18)              //P(E2 ∩ E1) = P(E1)P(E2|E1)

P(what we are looking for) = 1/6 + 9/36     = 5/12