If five dice are thrown simultaneously, what is the probability of getting the sum as seven?
Solution:Option(A) is correct
The following are two cases when the sum will be 7:
$7=1+1+1+1+3 \Rightarrow ^5C_1$ ways = 5
$7=1+1+1+2+2 \Rightarrow ^5C_2$ ways = 10
Total number of possible ways of throwing five dice \(=6^5\)
The required probability
Edit: Based on Poonam Pipaliya's comment, solution has been modified.
Error(s) Found !!!
Sudhanva Dixit (Jan 11'16 at 00:26)
How is $1+1+1+2+2=^5C_2$ ways?
It is not mathematically equal, it equals the number of ways three, 1's and two, 2's can be obtained from five dice.
Poonam Pipaliya (Nov 24'15 at 17:44)
in second line there should be $1+1+1+2+2=7$
Yes Poonam, you are right, corrected the solution.
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To write Maths use $ or $$ delimiters. (TeX)Ex: $ax^2+bx+c=0$.
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