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  • Three dice are thrown. The probability of getting a sum which is a perfect square isOption: 1 <img alt="\frac{2}{5}" src="https://entrancecorner.onco

Three dice are thrown. The probability of getting a sum which is a perfect square is

Option: 1

\frac{2}{5}


Option: 2

\frac{9}{20}


Option: 3

\frac{1}{4}


Option: 4

None of these


Answers (1)

best_answer

\mathrm{n(S)=6 \times 6 \times 6}. Clearly, the sum varies from 3 to 18 , and among these 4 , 9,16 are perfect squares.

The number of ways to get the sum 4

         = The number of integral solutions of \mathrm{x_1+x_2+x_3=4},
                                                             where \mathrm{1 \leq x_1 \leq 6,1 \leq x_2 \leq 6,1 \leq x_3 \leq 6}

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

Similarly, the number of ways to get the sum 9 

         \begin{aligned} & =\text { coefficient of x}^6 \text { in } \mathrm{\left(1-x^6\right)^3 \cdot(1-x)^{-3}} \\ \\& = \mathrm{-3 \times 1+{ }^8 C_6=28-3=25} \end{aligned}

The number of ways to get the sum 16

          \begin{aligned} & =\text { coefficient of x}^{13} \text { in } \mathrm{\left(1-x^6\right)^3 \cdot(1-x)^{-3}} \\ \\& =\text { coefficient of x}^{13} \text { in } \mathrm{\left(1-3 x^6+3 x^{12}-x^{18}\right)\left({ }^2 C_0+{ }^3 C_1 x+{ }^4 C_2 x^2+\ldots\right)} \\ \\& = \mathrm{{ }^{15} C_{13}-3 \times{ }^9 C_7+3 \times{ }^3 C_1=105-108+9=6} \end{aligned}

\mathrm{\therefore \quad n(E)=3+25+6=34 \text {. So, } P(E)=\frac{34}{6 \times 6 \times 6}=\frac{17}{108} \text {. }}

Posted by

vishal kumar

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