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Raj tossed 3 dices and their results are noted down then what is the probability that Raj gets 10?

I know the simple way to count the total favorable number of condition divide by (6^3). But it is time-consuming to solve in the examination to count one by one favorable condition.

So, solve by some short trick and give a generalized the formulae.

1 Answer

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If you think logically , the counting can be done in less than 1 minute but of course it requires practise. 

$First Observation:$ There will be at least one of the $4,5,6$ because even if you take $3$ three times, you cannot make a $10$.

Number of ways in which 4 can be involved:

$4,5,1$ total $3! = 6$ ways

$4,4,2$ total $3$ ways

$4,3,3$ total $3$ ways

Number of ways in which 5 can be involved:

$5,4,1$ has a $4$ so dont count

$5,3,2$ total $3! = 6$ ways

Number of ways in which 6 can be involved:

$6,3,1$  total $3! = 6$ ways

$6,2,2$ total $3$ ways

$Total count = 27$. Probability is $1/8$

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