GATE CSE 2012 | Question: 33

Question

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. $\dfrac{10}{21}$
• B. $\dfrac{5}{12}$
• C. $\dfrac{2}{3}$
• D. $\dfrac{1}{6}$

Comments

the question says , “if” 1,2,or 3 comes the die is rolled second time.”1,2 or 3” is not fixed to come on first roll.

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The question says 

we roll the first time .

Now if we get 1 ,2 or 3 we go for the second roll or else not .

What is the probability that the sum of values in the above event turn up to be at least 6? 

Now go and check @KUSHAGRA गुप्ता comment  it will be crystal clear.

 

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@Arjun sir @Hira Thakur sir here there is no info given that we can consider only one dice rolled and got 6 will also can be consider fabvourable case .

how can we know that ?

Answer 1

Total Probablity = 1/6 + 3/6{9/18} = 5/12

Answer 2

Probability that sum must be at least 6 $ = 1 – P(sum < 6)$

Case1: 

4, 5 are the only terms that are less than 6, and each has a probability of $1/6$, and hence $1/3$

Case 2: 

Now, when the value on die is $1, 2, 3 $, the total favourable outcomes are: 

$(1,1), (1, 2), (1, 3), (1, 4)$

$ (2, 1), (2, 2), (2, 3)$

$(3, 1), (3, 2),$

out of total 36 outcomes. 

So, $P(sum < 6)$ = $P(case 1) + P(case 2)$

$ = 7/12$ 

$P(sum >= 6) = 1 – P(sum < 6)$

$= 1 – 7/12$

$ = 5/12$

Answer 3

p(s>=6) = 1-p(s<6)

p(s<6)= p(4) + p(5) + p(1+2+3)

for 4 and 5 in first turn there is no next move so it is simply 1/6+ 1/6=1/3

so for 1 in first turn if 1,2,3,4 turn up in next move then sum will be less than 6 same for 2 (1,2,3) and for 3(1,2)

p(1+2+3)= 1/6 * (4/6) + 1/6 * (3/6)  + 1/6 *(2/6) = 1/4

p(s<6)= 1/3 + 1/4 = 7/12

p(s>=6) = 1- 7/12 = 5/12

Answer 4

getting 6 in first try is 1/6

then remain (1,5)(1,6)..(3,6) 9 element each has 1/36 probablity

so 1/6+9/36=5/12