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+49 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 $6$ ?

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

+96 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: $B$

0

IN THE PROBLEM HE SAID SUM MUST BE ATLEAT 6.....

THEN HOW DID U TAKE '6' AS ALONE SIR!!!!1 AS AN EVENT

THEN HOW DID U TAKE '6' AS ALONE SIR!!!!1 AS AN EVENT

+1

*If the value on the die is 1, 2, or 3, the die is rolled a second time.*

So, if there are 4,5 or 6 on the first roll, there will not be any second roll. So, {6} is taken as a favorable case.

0

@arjun sir @bikram sir

I perfectly got this approach.

this question is given under Bayes' theorem in Go pdf, but what would be the events look like if try to do it Bayes' theorem. I've tried like this

P(E1)= Rolling a dice. P(E2)=getting 1/2/3 P(E3)= getting sum >=6 in second roll. but not able to continue, will you please help here.

I perfectly got this approach.

this question is given under Bayes' theorem in Go pdf, but what would be the events look like if try to do it Bayes' theorem. I've tried like this

P(E1)= Rolling a dice. P(E2)=getting 1/2/3 P(E3)= getting sum >=6 in second roll. but not able to continue, will you please help here.

+3

0

@namita

becasue by the time u roll it the second time....the favourable case is the sum of the two values which should be >=6

becasue by the time u roll it the second time....the favourable case is the sum of the two values which should be >=6

+5

//This is what I feel.

**Viewpoint 1:**

Solution set: { 6, (1,5), (1,6) ......}

i.e. P(6 appeared on first throw) +

P(1 appeared on first throw** and** 5 appeared on second throw) +

P(1 appeared on first throw** and** 6 appeared on second throw) + ....................

= 1/6 + (1/6)(1/6) + (1/6)(1/6) + .....

= 1/6 + 9/36

= 5/12

**Viewpoint 2:**

P(......) = P(6 came on first throw) + P(sum>= 6 **and** 1,2,3 appeared in first throw)

= 1/6 + ????

P(1,2,3 appeared in first throw) = 1/2 //P(E1)

P(sum >= 6 | 1,2,3 appeared in first throw) = 9/18 //P(E2 | E1)

// Our new sample space is: { (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) }

// 9 favorable cases: {(1,5), (1,6), (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6) }

P(sum>= 6 **and** 1,2,3 appeared in first throw) = (1/2)(9/18) //P(E2 /\ E1) = P(E1)P(E2|E1)

P(what we are looking for) = 1/6 + 9/36 = 5/12

0

@namita

so the sample space of rolling dice once should be 3

and sample space of rolling dice twice 18

i suppose

so the sample space of rolling dice once should be 3

and sample space of rolling dice twice 18

i suppose

0

It is,

We can get six from(123) on first face

or 456(on first phase)

These are mutually exclusive and collectively exhaustive events.

Now we can expand them in bayee diagram and get the same result

0

Those asking why sample space is 1/36 for (1.5) and others.

See ,dices are always independent events.Now probability of getting 1 on first is 1/6 and 5 on second is 1/6

so independent events probability is multiplies =>$1/6*1/6 =1/36$

See ,dices are always independent events.Now probability of getting 1 on first is 1/6 and 5 on second is 1/6

so independent events probability is multiplies =>$1/6*1/6 =1/36$

0

What is Bayee diagram?

Will you please solve using Baye's Theorem formula.Because after reading your comment still i don't get clear picture of Baye's Theorem implementation.I find difficulty while using Baye's Theorem.

Will you please solve using Baye's Theorem formula.Because after reading your comment still i don't get clear picture of Baye's Theorem implementation.I find difficulty while using Baye's Theorem.

+2

@rahul

yes , thats in case if the first dice is either 1 ,2 ,3

but if the first dice is either 4,5,6 we dont throw second time

so its dependent event right ?

how can we take the same entire sample space

yes , thats in case if the first dice is either 1 ,2 ,3

but if the first dice is either 4,5,6 we dont throw second time

so its dependent event right ?

how can we take the same entire sample space

+2

THink like this?

What is probability of (1,5) ?

for 1 on first dice:- 6 choices 1/6

for 5 on second dice :- 6 choices. 1/6

Joint =1/36

Probability of throwing next time or not is dependent but probability of getting something when next time is throw is independent of what came previous time

Whether i throw or not is dependent on first result but if i throw,what i get is independent.

What is probability of (1,5) ?

for 1 on first dice:- 6 choices 1/6

for 5 on second dice :- 6 choices. 1/6

Joint =1/36

Probability of throwing next time or not is dependent but probability of getting something when next time is throw is independent of what came previous time

Whether i throw or not is dependent on first result but if i throw,what i get is independent.

0

@rahul got it :)

so only if it was events without replacement it actually affectes the sample space right ?

so only if it was events without replacement it actually affectes the sample space right ?

0

I still have this question that if we say that 1,2 or 5 has occured then the sample space

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

And we need to choose :

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

So the probability=9/18=1/2.

And then the answer should be 1/6+1/2=2/3.

I know its a silly question but what is wrong with my logic??

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

And we need to choose :

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

So the probability=9/18=1/2.

And then the answer should be 1/6+1/2=2/3.

I know its a silly question but what is wrong with my logic??

+26 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..**

0

The question is :-**What is the probability that the sum total of values that turn up is at least 6?**

so u may get 6 in first try also ..and it's Probability is 1/6.

+9

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!

+18 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**

+13 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.

+12 votes

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

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

+7 votes

**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

+5 votes

Let, Ei be the event that i shows(1<= i <= 6) when a die is rolled. Clearly ,P(Ei)=1/6

Let, A be the event that sum is at least 6.

we need to find , P(E6)+ P(E1 ∩ A) + P(E2 ∩ A) + P(E3 ∩ A)

= 1/6 + P(E1) P(A/E1) + P(E2) P(A/E2) + P(E3) P(A/E3)

=1/6 + 1/6 * 2/6 + 1/6 * 3/6 + 1/6 * 4/6 [Note : P(A/E1)=2/6 because if you get 1 after first roll, you have to get either 5 or 6 after second roll in order to make total sum at least 6. Similarly P(A/E2)=3/6 because if you get 2 after first roll,you can get 4 or 5 or 6 after second roll]

=1/6+1/6 * 9/6= 5/12

Let, A be the event that sum is at least 6.

we need to find , P(E6)+ P(E1 ∩ A) + P(E2 ∩ A) + P(E3 ∩ A)

= 1/6 + P(E1) P(A/E1) + P(E2) P(A/E2) + P(E3) P(A/E3)

=1/6 + 1/6 * 2/6 + 1/6 * 3/6 + 1/6 * 4/6 [Note : P(A/E1)=2/6 because if you get 1 after first roll, you have to get either 5 or 6 after second roll in order to make total sum at least 6. Similarly P(A/E2)=3/6 because if you get 2 after first roll,you can get 4 or 5 or 6 after second roll]

=1/6+1/6 * 9/6= 5/12

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