Probability question on a pair of dice

Question

I think the ans should be C, but the given answer is A. Anyone can explain please? 

Comments

you should believe yourself, answer is (C) only.

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Thanks

Answer 1

Ans will be (C)

A pair of dice rolled 3 times, like (a,b) ,(c,d) , (e,f)

So, total 6 throws

Now if sum of a pair is 10

that pair of throws could be (5,5) , (4,6) , (6,4)

Now, it is given among 3 pair of throws only one pair will show 10

Now, that pair we can choose among any one of (5,5),(4,6),(6,4)= so ³C₁ ways   (say (a,b) =³C₁)

Now for other two pairs ( say (c,d) , (e,f) ) we cannot choose among that previous three pairs

So, that can be   (say (c,d))

For 1 in one dice ,other dice can be 1,2,3,4,5,6 = 6 choice..................i

For 2  "     "    "           "    "       "     "   1,2,3,4,5,6 = 6 choice.................ii

For 3  "     "    "           "    "       "     "   1,2,3,4,5,6 = 6 choice..................iii

For 4  "     "    "           "    "       "     "   1,2,3,4,5 = 5 choice.....................iv

For 5  "     "    "           "    "       "     "   1,2,3,4,6 = 5 choice......................v

For 6  "     "    "           "    "       "     "   1,2,3,5,6 = 5 choice.....................vi

So, adding from (i) to (vi) we get 33 choices

So, for 1 pair we have 33 choices

Finally we can say for 3 pairs we have

                                   ³C_1* (³C₁ *33*33) / (36 * 36 * 36 )

                                   =121 / 576

      

Comments

@ Srestha

I think ,you have calculated probability for not showing sum '10'.You have not included favourable cases.

Your final answer should be multiplied by 3 corresponding to 3 favourable cases.

It should be (³C₁X3X33X33)/ (36X36X36) = 121/576

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The sum can be 10 in first throw (which you have considered), second or third. So, answer must be multiplied by 3.

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oh yes, thanks

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Also, you should not count cases like this. We have 6*6 = 36 total cases. So, just subtract 3 and get 33 :)

Answer 2

Option (C) is correct.

Let P(10) = probability of getting sum on pair of dices in single trial of throwing the two dices.

P(10 )  =  3/36

P^'(10) =  1-3/36 = 33/36

Using Binomial distribution, Probability of getting sum '10' exactly once in 3 trials of pair of dices =  

 ^{= 3}C₁ X P(10)¹X[P^'(10)]²  = 3x(3/36)x(33/36)² = 121 / 576