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85 votes
85 votes

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 $6$ ?

  1. $\dfrac{10}{21}$
  2. $\dfrac{5}{12}$
  3. $\dfrac{2}{3}$
  4. $\dfrac{1}{6}$
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16 Answers

12 votes
12 votes
P(getting a 1)=1/6 with 1 you can form 6,7 so p(sum>=6)=2/6

P(getting a 2)=1/6 with 2 you can form 6,7,8 so p(sum>=6)=3/6

P(getting a 3)=1/6 with 3 you can form 6,7,8,9 so p(sum>=6)=4/6

now every time we are going to form a number we require a particular number whose probability is 1/6

 

so total probability =1/6(2/6+3/6+4/6+1)=15/36=5/12
8 votes
8 votes

P(getting sum is atleast 6) = P(getting 6 in the first roll of die ) OR P(getting 1,2 or 3 in first AND getting no  according  to result of first roll which gives sum atleast 6)

P(sum>=6) = 1/6 + (1/2*9/18)

                         =  5/12

6 votes
6 votes
Let, Ei be the event that i shows(1<= i <= 6)  when a die is rolled.  Clearly ,P(Ei)=1/6

Let, A be the event that sum is at least 6.

we need to find , P(E6)+ P(E1 ∩ A) + P(E2 ∩ A) + P(E3 ∩ A)

= 1/6 + P(E1) P(A/E1) + P(E2) P(A/E2) + P(E3) P(A/E3)

=1/6 + 1/6 * 2/6 + 1/6 * 3/6 + 1/6 * 4/6  [Note : P(A/E1)=2/6 because  if you get 1 after first roll, you have to get either 5 or 6 after second roll in order  to make total sum at least 6. Similarly P(A/E2)=3/6 because  if you get 2 after first roll,you can get 4 or 5 or 6 after second roll]

=1/6+1/6 * 9/6= 5/12
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