Question

A child throws 2 fair dice. If the numbers showing are unequal, he adds them together to get his final score. On the other hand, if the numbers showings are equal, he throws 2 more dice & adds all 4 numbers showing to get his final score. The probability that his final score is 6 is?

A. $\dfrac{{145}}{{1296}}$

B.$\dfrac{{146}}{{1296}}$

C.$\dfrac{{147}}{{1296}}$

D.$\dfrac{{148}}{{1296}}$

Answer 1

Hint : In this question first we will find the probability of throwing dice when the number on two dice are unequal and then we will find the probability of throwing two dice where numbers are equal and they add up to 6.

Complete step-by-step answer :

We know the total numbers of outcomes when two dices are thrown $= 36$

Now when dices are thrown and the number showing on the two dices is unequal then we will add the number showing on the dices together and the total number of outcomes of getting unequal numbers which adds up to 6= {(1,5),(2,4),(4,2),(5,1)}, so the numbers of outcome $= 4$

Hence the probability of getting unequal numbers

$\Rightarrow P\left( {UE} \right) = \dfrac{4}{{36}} = \dfrac{1}{9}$

Now it is said when the numbers on the two dices are equal then the child throws 2 more dices and adds all the 4 numbers showing on the dices and select the numbers which adds to 6

Hence the numbers of outcomes of getting equal numbers

On first two dices (1,1) where two numbers are equal we throw two more dices to get the sum of 6, so the second two dices will be {(1,3),(2,2),(3,1)}

Now if first two dices (2,2) where two numbers are equal we throw two more dices to get the sum of 6, so the second two dices will be {(1,1)}

Hence the probability of getting equal numbers will be

$\Rightarrow P\left( E \right) = \left( {\dfrac{1}{{36}} \times \dfrac{3}{{36}}} \right) + \left( {\dfrac{1}{{36}} \times \dfrac{1}{{36}}} \right) = \dfrac{4}{{1296}}$

Therefore the probability that his final score is 6

$P\left( E \right) + P\left( {UE} \right) = \dfrac{4}{{1296}} + \dfrac{4}{{36}} = \dfrac{{4 + \left( {4 \times 36} \right)}}{{1296}} = \dfrac{{148}}{{1296}}$

So, the correct answer is “Option D”.

Note : The total outcomes when two dices are thrown {(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), (4,1),(4,2),(4,3),(4,4),(4,5),(4,6), (5,1),(5,2),(5,3),(5,4),(5,5),(5,6), (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)} is 36.