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 showing 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?

\frac{145}{1296}

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\frac{146}{1296}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{147}{1296}

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\frac{148}{1296}

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Solution

The correct option is D $\frac{148}{1296}$
Total outcomes=666*6=1296
cases when they are unequal : (1,5)(2,4)(4,2)(5,1)
cases when they are equal :(1,1,2,2)(1,1,1,3)(1,1,3,1)(2,2,1,1)
So, the possibilities after the first two throws are:
4/36 we have a total of 6 without a double and we stop.
26/36 we don't have a total of 6 or a double and we stop.
6/36 we have a double and must continue. We need to look more carefully.
1/36 we have a double 1, we throw 2 more dice and have a 3/36 chance of getting a total of 6.
1/36 we have double 2, we throw 2 more and have a 1/36 chance of getting 6.
4/36 we have a double 3 or more, we throw again but we can't get 6.
So, overall a successful outcome has probability of 4/36+1/36×3/36+1/36×1/36=148/1296

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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 showing 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 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 showing 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?