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It is given that from a box containing $15$ blue balls and $45$ black balls, $2$ balls are selected randomly without replacement.

Probability that, the first selected is a blue ball and the second selected is a black ball

$\qquad = \dfrac{^{15}C_{1}}{60} \times \dfrac{^{45}C_{1}}{59}$

$\qquad = \dfrac{15}{60} \times \dfrac{45}{59} = \dfrac{45}{236}.$

So, the correct answer is $(B).$

Probability that, the first selected is a blue ball and the second selected is a black ball

$\qquad = \dfrac{^{15}C_{1}}{60} \times \dfrac{^{45}C_{1}}{59}$

$\qquad = \dfrac{15}{60} \times \dfrac{45}{59} = \dfrac{45}{236}.$

So, the correct answer is $(B).$