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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. $\dfrac{1}{6}$
edited | 5.8k views
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Hello sir, why we have taken sample space 36 although we not throw dice again when we got 4,5,6 .

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: $B$
answered by Veteran (396k points)
edited ago
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in favorable cases why  6 is considered ?
+3
great explanation @Arjun.
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IN THE PROBLEM HE SAID SUM MUST BE ATLEAT 6.....

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

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@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.
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beautiful explanation @arjun sir :)
+1
This question is not based on Baye's Theorem.
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Why have we taken sample space 36 although we have not thrown the dice again when we got 4,5,6 ??
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Why the probability of selecting the mutually exclusive events is not considered?

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@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
+2

@A_i_$_h That is my point. We are never going to roll it for the second time if we get 4,5 ... i.e. (4,1), (4,2) ... etc these cases will never be there. So there are less than 36 cases in our sample space of second time roll. +2 //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 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$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. +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 +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. 0 @rahul got it :) so only if it was events without replacement it actually affectes the sample space right ? 0 whenever dependency comes into the picture it affects sample space. 0 @Higgs well explained . 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?? 0 Let E1= 1,2 or 3 occurred. P(E1)=1/2 E2=(sum>=6) ={(1,5),(1,6),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6)} So P(E2)=9/36 (independent on E1) P(getting 6 in first throw)=1/6 and then the answer should be 1/6 + 9/36=5/12 +1 How (9/18)/(1/2) is 9/36???? it should be 0.5/0/0.5=1 0 @Prashank thanks... i will update it +24 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... answered by Boss (23.6k points) edited +1 @rajesh plzz elaborate how 24 ways ? 0 Thanks @BILLY and Plz see the edited ans. 0 IN THE FINAL STEP..... WHY DID U ADD 1*(1/6)? PLEASE EXPLAIN ME ELABORTE IN MANNER 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. 0 NIce:)) +8 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! +13 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 answered by Loyal (7.5k points) +10 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 answered by Boss (14.5k points) +1 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 +1 No , its ok . last one is added for anything that is not 1 or 2 or 3 , right . +8 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. answered by Junior (969 points) +6 votes 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 answered by (95 points) 0 @Rajesh,1/6 coz 6 can occur on either of d two dice.Isn't it? 0 @Devshree, 1/6 because 6 can come first time when we roll the dice,then we don't require to roll it again +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 answered by Active (2.4k points) +1 Using total probability !!! ... 0 Very good ans 0 I'm missing the term$P(E_6)\$.  ☺
Total Probablity = 1/6 + 3/6{9/18} = 5/12
answered by Active (2.7k points)
p(s>=6) = 1-p(s<6)

p(s<6)= p(4) + p(5) + p(1+2+3)

for 4 and 5 in first turn there is no next move so it is simply 1/6+ 1/6=1/3

so for 1 in first turn if 1,2,3,4 turn up in next move then sum will be less than 6 same for 2 (1,2,3) and for 3(1,2)

p(1+2+3)= 1/6 * (4/6) + 1/6 * (3/6)  + 1/6 *(2/6) = 1/4

p(s<6)= 1/3 + 1/4 = 7/12

p(s>=6) = 1- 7/12 = 5/12
answered by (83 points)

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