Question

A box B₁ contains 1 white ball, 3 red balls and 2 black balls. Another box B₂ contains 2 white balls, 3 red balls and 4 black balls. A third box B₃ contains 3 white balls, 4 red balls and 5 black balls.

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Linked Question 1

If 1 ball is drawn from each of the boxes B₁, B₂ and B₃, the probability that all 3 drawn balls are of the same colour is

$\frac{90}{648}$

$\frac{566}{648}$

$\frac{82}{648}$

$\frac{558}{648}$

(i) $Probability = \frac{\frac{1}{3} \times (\frac{^{2} C_{1} \times^{3} C_{1}}{^{9} C_{2}})}{\frac{1}{3} \times [\frac{^{1} C_{1} \times^{3} C_{1}}{^{6} C_{2}} + \frac{^{2} C_{1} \times^{3} C_{1}}{^{9} C_{2}} + \frac{^{3} C_{1} \times^{4} C_{1}}{^{12} C_{2}}]} = \frac{\frac{6}{36}}{\frac{13}{15} + \frac{6}{30} + \frac{12}{66}} = \frac{\frac{1}{6}}{\frac{1}{5} + \frac{1}{6} + \frac{2}{11}} = \frac{55}{181} .$ Ans.

(ii) $Probability = (\frac{1}{6} \times \frac{2}{9} \times \frac{3}{12}) + (\frac{3}{6} \times \frac{3}{9} \times \frac{4}{12}) = \frac{6 + 36 + 40}{72 \times 9} = \frac{82}{648} .$ Ans.

Linked Question 2

If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box B₂ is

$\frac{65}{181}$

$\frac{55}{181}$

$\frac{126}{181}$

$\frac{116}{181}$

(ii) $Probability = (\frac{1}{6} \times \frac{2}{9} \times \frac{3}{12}) + (\frac{3}{6} \times \frac{3}{9} \times \frac{4}{12}) = \frac{6 + 36 + 40}{72 \times 9} = \frac{82}{648} .$ Ans.