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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.

Here our sample space consists of $3 + 3\times 6 = 21$ events- $(4), (5), (6), (1,1), (1,2)\ldots(3,6).$

Favorable cases $=(6), (1,5), (1,6), (2, 4), (2, 5), (2, 6), (3, 3), (3,4), (3,5), (3,6).$

Required Probability $= \dfrac {\text{No. of favorable cases}}{\text{Total cases }}= \dfrac{10}{21}$

But this is wrong way of doing. Because due to $2$ tosses for some and $1$ for some, individual probabilities are not the same. i.e., while $(6)$ has $\dfrac{1}{6}$ probability of occurrence, $(1,5)$ has only $\dfrac{1}{36}$ probability. So, our required probability

$\Rightarrow \dfrac{1}{6} + \left(9 \times \dfrac{1}{36}\right) =\dfrac{5}{12}.$

Correct Answer: $B$
by

thatswhy we are adding the case of 6 as well.

correct me if i am wrong.

thanks @Arjun Sir
why (9 * 1/36) and not (9 * 1/18)?

because for the outcomes

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

the total outcomes are 18 and not 36!

Following is the wrong approach which I had done initially and max. people think like this...So, we should know this(thanks @BILLY for identifying the mistake...) ///The last img is the correct approach..

### And the correct approach goes like below...

NIce:))

So, this one is such intresting question:

If (1 or 2 or 3) appear than dice rolled second time and some cases are favorable, but if (4 or 5 or 6) appear than dice don't rolled out second time!

And now, the twist comes, they said total sum will be at least 6(it dosen't matter that you get it by 1 roll of dice or 2 roll of dice) so if 6 comes than we stop for rolling but it is our satisfying condition!

Such a great question!

edited
shouldn’t it be 1/18 instead of 1/36? Because total outcomes will only be calculated if the number on the first throw comes 1 or 2 or 3?

Question is simple if you just think little hard.

Given that when {1,2,3} occurs then rolled the die 2nd time, and when {4,5,6} is occur just stop.By this, we can say that max-sum is 9 (3,6)and min-sum is 2 (1,1),

Let P(x) means prob of value x on single die and P(x,y) means prob of value x on 1st die and y on the 2nd die (in that order bcoz order is matter). okay now.

P(sum >=6) = P(sum=6) +P(sum=7) +P(sum=8) +P(sum=9)         ////just find all the values

P(sum=6) = P(6) +P(1,5)+P(2,4)+P(3,3) = 3* 1/36  +1/6 = 1/4

P(sum=7) = P(1,6) +P(2,5)+P(3,4)= 3* 1/36  = 1/12

P(sum=8) = P(2,6) +P(3,5)= 2* 1/36  = 1/18

P(sum=9) = P(3,6)= 1/36

so now,

P(sum >=6) =  1/4 +1/12​​​​​​​ +1/18​​​​​​​ +1/36  =15 /36 =5 /12

## The correct answer is (B) 5/12

by

We can solve this question by breaking it into two parts which is:

1) .Die is rolled only once i.e., either (4,5,6) comes.

2). Die is rolled twice i.e., (1,2,3) comes in first roll.

This is how I solved, just missed 6 in the first case, as I didn't think about that case.
why did we took 1/36 in the 6 case as we are not rolling dice if outcome is 4,5 or 6?

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