BIG CONFUSION _PROBABILITY

Question

1)If two fair dice is rolled ,find the probability that sum is 6 when 1st die is 4   

2)find the probability of getting a 2 in the second dice if first die is 4

 

can we say these two are same question beacuse for 1st one answer is 1/5 for second one it is coming 1/6

Reason: in 1st one we can not consider the cases where 6 comes in 1st die in sample space,coz no chance of sum coming 6 then

Answer 1

1. If two fair dice are rolled, find the probability that sum is 6 when 1st die is 4.
2. Find the probability of getting a 2 in the second dice if first die is 4.

These two questions are exactly the same and answer is $\frac{1}{6}$.

To make it clear lets do another question

1. If two fair dice are rolled, and their sum is 6, find the probability that the first one is 4.

We have fair dices. So, all the possible occurences have equal chance.

Probability of getting a sum 6 $= \frac{\text{Favorable Cases}}{\text{Total Cases}}$

$ = \frac{5}{36}$

as favorable cases are $(1,5),  (2,4),  (3,3), (4,2), (5,1)$.

Now, probability of first die being 4 and sum being 6 $ = \frac{1}{6} \times\frac{1}{6} = \frac{1}{36},$

as here the only option for the second die is to show 2.

Now as per Bayes' theorem, probability of first die being 4 given that sum is 6,

$ P(A \mid B)= \frac{P(AB)}{P(B)} = \frac{\frac{1}{36}} {\frac{5}{36}} = \frac{1}{5}.$

Comments

Please ask your doubt in proper sentence - as I told one word missing changes the question.

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OKAY,Now the entire scenario is cleared to me

If the question is to find the probability of sum 6 when first is 4 then {(4,1)...(4,6)} so answer is 1/6

but if it is to find probability of first being 4 when sum is 6 is {(3,3)(4,2)(2,4)(1,5)(5,1)}  so 1/5

thanks arjun sir for clearing

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yes. But you have to be careful with that approach for conditional probability - that is counting the items in sample space. Here it works as all the 5 outcomes have same probabilities but may not be with other cases.

Answer 2

1)the sample spaces are when Sum is  6={(1,5),(2,4),(3,3),(4,2),(5,1)}

but we want the case when 1st die is 4 i.e. (4,2)

So, probability 1/5

 

2)1st die is 4 in the cases {(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)}

But we want 2nd die to be 2 i.e.(4,2)

So, probability = 1/6

Comments

YES! @srestha i did the same using this method and conditional ptopbability also

but htse question is almost very  same but what a difference ... but if it were that sum is 7 and first is 4 and second is 3 and first is 4.. then they will be same question...

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yes it will change accordingly

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You are solving the wrong question. To get sum 6 given that first die is 4, sample space is {1,2,3,4,5,6} and favorable is {2}.

Answer 3

For 1st ,

A=sum is 6

B=1st dice is 4

P(A)=5/36        

P(B)=6/36

probability of having sum 6 and 1st die=4  ie P(AB)=1/36

probability of getting a 6 when 1 st dice is 4 = P(A/B)=P(AB)/P(B)

         = 1/36/6/36 = 1/6

for 2nd,

C=2 in 2nd dice

D= 4 in 1st dice

P(D)=6/36

P(CD)=1/36

P(C/D)=1/36/6/36 = 1/6

Answer 4

Your 1st question is -- If two fair dice is rolled ,find the probability that sum is 6 when exactly one die show 4... ryt...?

In this case 1st die is 4

2nd die can have any no except 4 ..(Since no 2 die can have 4)

therefore, out of 6 possible outcomes u can have only 5 outcomes

ans= 1/5

ok...??

Comments

no not one die... first die for exact