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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. $\dfrac{1}{6}$

How $6$ is valid?

On first dice only $1,2,3$ can come.
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.

I followed this approach​​​​​

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(t6)=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

by

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

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