Choose the correct answer from the given four options.The probability that exactly two of the three balls were red, the first ball being red, is:

Q: Choose the correct answer from the given four options.The probability that exactly two of the three balls were red, the first ball being red, is:

Let $E_1 =$ Event that first ball being red And $E_2 =$ Event that exactly two of three balls being red $\therefore\text{P}(\text{E}_1)=\text{P}_{\text{R}}\cdot\text{P}_{\text{R}}\cdot\text{P}_{\text{R}}+\text{P}_{\text{R}}\cdot\text{P}_{\text{R}}\cdot\text{P}{_\bar{\text{R}}}+\text{P}_{\text{R}}\cdot\text{P}{_\bar{\text{R}}}\cdot\text{P}_{\text{R}}+\text{P}_{\text{R}}\cdot\text{P}{_\bar{\text{R}}}\cdot\text{P}{_\bar{\text{R}}}$ $=\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}+\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}+\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{4}{6}+\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{2}{6}$ $=\frac{60+60+60+30}{336}=\frac{210}{336}$ $\text{P}(\text{E}_1\cap\text{E}_2)=\text{P}{_\text{R}}\cdot\text{P}{_\bar{\text{R}}}\cdot\text{P}{_\text{R}}+\text{P}{_\text{R}}\cdot\text{P}{_\text{R}}\cdot\text{P}{_\bar{\text{R}}}$ $=\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{4}{6}+\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}=\frac{120}{336}$ $\therefore\text{P}\Big(\frac{\text{E}_2}{\text{E}_1}\Big)=\frac{\text{P}(\text{E}_1\cap\text{E}_2)}{\text{P}(\text{E}_1)}$ $=\frac{\frac{120}{336}}{\frac{210}{336}}=\frac{4}{7}$

M.C.Q (1 Marks)

GSEB - English Medium

Choose the correct answer from the given four options.The probability that exactly two of the three balls were red, the first ball being red, is:

A$\frac{1}{3}$
B$\frac{4}{7}$
C$\frac{15}{28}$
D$\frac{5}{28}$

STD 12 Science
Maths
Probability
Exemplar

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