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Question

Answers

A. $\dfrac{{47}}{{66}}$

B. $\dfrac{{23}}{{33}}$

C. $\dfrac{{47}}{{132}}$

D. $\dfrac{{47}}{{33}}$

E. $\dfrac{{70}}{{33}}$

Answer

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$P(A) = $ Total number of the favourable outcomes / Total number of the outcomes

$ = 12{\rm{ balls}}$

${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$

${}^{12}{C_2} = \dfrac{{12!}}{{2!(12 - 2)!}}$

$\begin{array}{l}

{}^{12}{C_2} = \dfrac{{12 \times 11 \times 10!}}{{2(10!)}}\\

{}^{12}{C_2} = 66

\end{array}$

Similarly,

$\begin{array}{l}

{}^3{C_2} = 3\\

{}^4{C_2} = 6\\

{}^5{C_2} = 10

\end{array}$

Favourable number of cases where balls are of different colours

$ = {}^{12}{C_2} - ({}^3{C_2} + {}^4{C_2} + {}^5{C_2})$

$\begin{array}{l}

= 66 - (3 + 6 + 10)\\

= 66 - 19\\

= 47

\end{array}$

The probability that balls are of the different colours,

$\begin{array}{l}

= \dfrac{{Favourable{\rm{ Cases}}}}{{Total{\rm{ number of cases}}}}\\

= \dfrac{{47}}{{66}}

\end{array}$

Therefore, the required solution- The probability that balls are of different colours is $ = \dfrac{{47}}{{66}}$

Hence, from the given multiple choices option A is the correct answer.