Probability Question 1
Question 1 - 2024 (01 Feb Shift 1)
A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is:
(1) $\frac{2}{5}$
(2) $\frac{2}{7}$
(3) $\frac{1}{7}$
(4) $\frac{1}{5}$
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Answer (2)
Solution
$\mathrm{P}(4 \mathrm{~W} 4 \mathrm{~B} / 2 \mathrm{~W} 2 \mathrm{~B})=$
$P(4 W 4 B) \times P(2 W 2 B / 4 W 4 B)$
$\overline{P(2 W 6 B) \times P(2 W 2 B / 2 W 6 B)+P(3 W 5 B) \times P(2 W 2 B / 3 W 5 B)}$
$+\ldots \ldots \ldots+P(6 W 2 B) \times P(2 W 2 B / 6 W 2 B)$
$=\frac{\frac{1}{5} \times \frac{{ }^{4} \mathrm{C}{2} \times{ }^{4} \mathrm{C}{2}}{{ }^{8} \mathrm{C}{4}}}{\frac{1}{5} \times \frac{{ }^{2} \mathrm{C}{2} \times{ }^{6} \mathrm{C}{2}}{{ }^{8} \mathrm{C}{4}}+\frac{1}{5} \times \frac{{ }^{3} \mathrm{C}{2} \times{ }^{5} \mathrm{C}{2}}{{ }^{8} \mathrm{C}{4}}+\ldots+\frac{1}{5} \times \frac{{ }^{6} \mathrm{C}{2} \times{ }^{2} \mathrm{C}{2}}{{ }^{8} \mathrm{C}{4}}}$
$=\frac{2}{7}$
