GATE CSE 2012 | Question: 33

Question

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 $6$ ?

• A. $\dfrac{10}{21}$
• B. $\dfrac{5}{12}$
• C. $\dfrac{2}{3}$
• D. $\dfrac{1}{6}$

Comments

 Easiest Method using Tree Diagram

 

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see in dice 6 possible out come here question have asked sum need to be atleast 6 means 6 or more than 6

in the cases are if you got 1,2,3 u have a chance to roll again so

case 1: first time we roll we got 6

out of 6 one possibility so 1/6

case 2 : in any other way a single time we cant get more than 6 so there are 3 ways

1. if we get 1 we have a chance in that we need to get {5,6}

then sum will be more than 6 so 1/6 *2/6

2. if we get 2 we have possibility to get sum >= 6 are {4,5,6}

1/6*3/6

3. for 3 we can have possiblity {3,4,5,6}

1/6*4/6

so sum of all these 2/36+3/36+4/36 = 9/36 =1/4

total probability 1/4+1/6 =0.417 5/12 so ans is 5/12

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You need to understand the sample space here : It is given that is 1,2 or 3 comes then the die is thrown again now Sample space will look like:

{ (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)

4 , 5 , 6

}

now probability distribution is not equally likely because

say from the sample space 4 happens then it is 1/6 [ A single throw]

and say if (3,2) happens then its 1/36 [ Two throws ]

so yes they are different probabilities.  You can verify that the individual sample points sum up to 1, the total probability.

 

Now if 6 comes then = 1/6

if 3 comes then in second throw 3,4,5,6 should come = 1/6 * 4/6

for 2 : 1/6 * 3/6

for 1 : 1/6 * 2/6

 

add all.

Answer 1

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: $B$

Comments

-> On first roll if die shows <=3 then we roll again or else we don't

-> Ok, Now the number of Favourable cases for sum >= 6
      
    (1,5), (1,6), (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6) and (6) [ Why 6?.
     

     Since it alone also satifies condition sum>=6
    

     Sum of  Probabilities = 1/6 * 2/6 + 1/6*3/6 + 1/6*4/6 + 1/6 = 5/12

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anyone please answer why those 9 cases have 1/36 probability.

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I do that 1st mistake.

Answer 2

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...

Comments

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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thanks for  your kind efforts

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i have the same question bro...

Answer 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

Comments

Niceeee

Answer 4

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.

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?

Answer 5

P(getting a 1)=1/6 with 1 you can form 6,7 so p(sum>=6)=2/6

P(getting a 2)=1/6 with 2 you can form 6,7,8 so p(sum>=6)=3/6

P(getting a 3)=1/6 with 3 you can form 6,7,8,9 so p(sum>=6)=4/6

now every time we are going to form a number we require a particular number whose probability is 1/6

 

so total probability =1/6(2/6+3/6+4/6+1)=15/36=5/12

Comments

please mention bayes' theorem and the fact that these events are independent

also mention the case where you get 6 on you first roll

your answer is correct but explaination is incomplete

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No , its ok .

 

last one is added for anything that is not 1 or 2 or 3 , right .

Answer 6

P(getting sum is atleast 6) = P(getting 6 in the first roll of die ) OR P(getting 1,2 or 3 in first AND getting no  according  to result of first roll which gives sum atleast 6)

P(sum>=6) = 1/6 + (1/2*9/18)

                         =  5/12

Comments

@Rajesh,1/6 coz 6 can occur on either of d two dice.Isn't it?

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@Devshree,

1/6 because 6 can come first time when we roll the dice,then we don't require to roll it again