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

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69 votes

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

150 votes

Best answer

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: $B$

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: $B$

33 votes

**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..**

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!

11

22 votes

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**