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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 | 5.9k 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: $B$
edited
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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.
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@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
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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.
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@rahul got it :)

so only if it was events without replacement it actually affectes the sample space right ?
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whenever dependency comes into the picture it affects sample space.
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@Higgs well explained .
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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??
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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

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How (9/18)/(1/2) is 9/36???? it should be 0.5/0/0.5=1
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@Prashank thanks...

i will update it

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
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@rajesh plzz elaborate how 24 ways ?
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Thanks @BILLY and Plz see the edited ans.
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IN THE FINAL STEP.....

PLEASE EXPLAIN ME ELABORTE IN MANNER
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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.

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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!

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

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

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

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No , its ok .

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

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.

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

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@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
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
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Using total probability !!! ...
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Very good ans
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I'm missing the term $P(E_6)$.  ☺
Total Probablity = 1/6 + 3/6{9/18} = 5/12
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

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