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

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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 $6$ ?

- $\dfrac{10}{21}$
- $\dfrac{5}{12}$
- $\dfrac{2}{3}$
- $\dfrac{1}{6}$

2

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.

0

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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)**