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.

---

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.

 

---

@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

I followed this approach​​​​​ 

Answer 2

I solved with total probability theorem :

P(e)= p(sum>=6),

P(e)=can be divided into 2 mutually exclusive sets :

1. sum>=6  given twice throwi.e when 1st o/p(1,2,3)=p(t∩6)=9/36
2. sum>=6 given no 2nd throwi.e when 1st o/p is 4,5,6=p(o∩6)=1/6

p(e)=p(o∩6)+p(t∩6)= 1/6+9/36

Answer 3

You have got two ways of getting sum atleast 6

You get 6 in the first toss    (OR)    You get 1, 2 and 3 in the first toss and then get sum 6 with the second throw(this is simultaneous)…...eq1

P( getting 6 in the first toss ) =1/6

P(getting 1,2 and 3) =3/6=½
P(getting sum 6 in the second throw when 1,2 or 3 is thrown ) =>
favourable cases =(1,5) (1,6) (2,4) (2,5) (2,6) (3,3) (3,4) (3,5) (3,6) =9 nos
sample space possible when 1, 2 or 3 is thrown= (1,1) (1,2)…...(3,3)=18 nos

Therefore eq 1 can be evaluated as,
1/6+(1/2)*(9/18)= 5/12

Correct answer is (B)


 

Answer 4

Here its mention that we would roll the dice for another time if we got face value of 1, 2 and 3.

So, let us consider a case where the face value be : [1 , 1 , 1 , 1 , 1 , 1] (for upto min 6 times the dice shall roll)

Similarly for : [2 , 2 , 2] (3 times the dice rolled) , [3 , 3] (2 times the dice rolled) and [1, 2, 3] (1 time the dice will roll.

Above 4 cases satisfies the condition the face value is atleast 6.

Now, considering another favoured conditions where the face value can also be atleast 6:

[1, 5] , [1, 6] 

[2, 4] , [2, 5] , [2, 6]

[3, 3](already counted above so don’t repeat) , [3, 4] , [ 3, 5] , [3, 6]

[4] , [5] , [6]

 

So, now favourable conditions are : $\frac{(4 + 11)}{36}$ = $\frac{15}{36}$ = $\frac{5}{12}$