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# GATE | GATE CS 2012 | Question 31

• Difficulty Level : Medium
• Last Updated : 22 Mar, 2021

Suppose a fair six-sided die is rolled once. If the value on the die is 1, 2, or 3, the die is rolled aÂ 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 âˆ© E1) = P(E1)P(E2|E1)

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