Question

An urn contains 3 white, 4 red and 5 black balls. Two balls are drawn one by one without replacement. What is the probability that at least one ball is black?

Answer 1

Consider the given events.
A = A white or red ball in the first draw
B = A white or red ball in the second draw

$$\begin{gathered} \mathrm{Now}, \\ P\left(A\right)=\frac{7}{12} \\ P\left(B/A\right)=\frac{6}{11} \\ \therefore P\left(A\cap B\right)=P\left(A\right)\times P\left(B/A\right) \\ =\frac{7}{12}\times \frac{6}{11} \\ =\frac{7}{22} \\ \therefore \mathrm{Required}\mathrm{probability}=1-P\left(A\cap B\right) \\ =1-\frac{7}{22} \\ =\frac{15}{22} \end{gathered}$$