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69 votes
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 $6$ ?

  1. $\dfrac{10}{21}$
  2. $\dfrac{5}{12}$
  3. $\dfrac{2}{3}$
  4. $\dfrac{1}{6}$
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4 Comments

How $6$ is valid?

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

 

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15 Answers

150 votes
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: $B$
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4 Comments

  thatswhy we are adding the case of 6 as well.

correct me if i am wrong.

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thanks @Arjun Sir
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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!
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33 votes
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..

And the correct approach goes like below...

edited by

4 Comments

NIce:))
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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!

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edited by
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?
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22 votes
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

The correct answer is (B) 5/12

17 votes
17 votes

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.

2 Comments

This is how I solved, just missed 6 in the first case, as I didn't think about that case.
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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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Answer:

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