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A bag contains $3$ red, $4$ white and $5$ blue balls. If two balls are drawn at random, then the probability that they are of different colours, isA. $\dfrac{{47}}{{66}}$ B. $\dfrac{{23}}{{33}}$ C. $\dfrac{{47}}{{132}}$ D. $\dfrac{{47}}{{33}}$ E. $\dfrac{{70}}{{33}}$

Last updated date: 20th Jun 2024
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Hint: Probability of any given event is equal to the ratio of the favourable outcomes with the total number of the outcomes. Probability is the state of being probable and the extent to which something is likely to happen in the particular situations.
$P(A) =$ Total number of the favourable outcomes / Total number of the outcomes

Complete step by step solution:Total number of observations $= 3{\rm{ red + 4 white + 5 blue balls}}$
$= 12{\rm{ balls}}$
${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
${}^{12}{C_2} = \dfrac{{12!}}{{2!(12 - 2)!}}$
$\begin{array}{l} {}^{12}{C_2} = \dfrac{{12 \times 11 \times 10!}}{{2(10!)}}\\ {}^{12}{C_2} = 66 \end{array}$
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.

Note: Combinations are used if certain objects are to be arranged in such a way that the order of objects is not important whereas Permutation is an ordered combination- an act of arranging the objects or numbers in the specific order.