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  • The probability of getting a sum of 12 in four throws of an ordinary dice isOption: 1 <img alt="\frac{1}{6}\left(\frac{5}{6}\right)^3" src="https://e

The probability of getting a sum of 12 in four throws of an ordinary dice is

Option: 1

\frac{1}{6}\left(\frac{5}{6}\right)^3


Option: 2

\left(\frac{5}{6}\right)^4


Option: 3

\frac{1}{36}\left(\frac{5}{6}\right)^3


Option: 4

None of these


Answers (1)

best_answer

\mathrm{n(S)=6 \times 6 \times 6 \times 6}

\mathrm{n(E)} = the number of integral solutions of \mathrm{x_1+x_2+x_3+x_4=12,}

                                                                              where \mathrm{1 \leq x_1 \leq 6, \ldots, 1 \leq x_4 \leq 6}

           =\text { coefficient of x}^{12} \text { in } \mathrm{\left(x+x^2+\ldots+x^6\right)^4}

            \begin{aligned} & =\text { coefficient of x}^8 \text { in }\mathrm{\left(\frac{1-x^6}{1-x}\right)^4} \\ \\& =\text { coefficient of x}^8 \text { in } \mathrm{\left(1-x^6\right)^4 \cdot\left({ }^3 C_0+{ }^4 C_1 x+{ }^5 C_2 x^2+\ldots\right)} \\ \\& =\mathrm{{ }^{11} C_8-4 \cdot{ }^5 C_2=125} \end{aligned}

\mathrm{\therefore \quad P(E)=\frac{125}{6 \times 6 \times 6 \times 6} .}

Posted by

rishi.raj

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