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
Answer: (B)
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
Correct Answer: B
Quiz of this Question