Question

(i) 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?
(ii) A bag contains 4 white, 7 black and 5 red balls. Three balls are drawn one after the other without replacement. Find the probability that the balls drawn are white, black and red respectively.

Answer 1

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

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

(ii) Consider the given events.
A = A white ball in the first draw
B = A black ball in the second draw
C = A red ball in the third draw

$$\begin{gathered} \mathrm{Now}, \\ P\left(A\right)=\frac{4}{16}=\frac{1}{4} \\ P\left(B/A\right)=\frac{7}{15} \\ P\left(C/A\cap B\right)=\frac{5}{14} \\ \therefore \mathrm{Required}\mathrm{probability}=P\left(A\cap B\cap C\right)=P\left(A\right)\times P\left(B/A\right)\times P\left(C/A\cap B\right) \\ =\frac{1}{4}\times \frac{7}{15}\times \frac{5}{14} \\ =\frac{1}{24} \end{gathered}$$