Choose the correct answer from the given four options. The probab… - Vidyadip

MCQ

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
$\frac{5}{28}$

✓

Answer

$\frac{4}{7}$

Solution:

Let E₁= Event that first ball being red

And E₂ = Event that exactly two of three balls being red

$\therefore\text{P}(\text{E}_1)=\text{P}_{\text{R}}\cdot\text{P}_{\text{R}}\cdot\text{P}_{\text{R}}+\text{P}_{\text{R}}\cdot\text{P}_{\text{R}}\cdot\text{P}{_\bar{\text{R}}}+\text{P}_{\text{R}}\cdot\text{P}{_\bar{\text{R}}}\cdot\text{P}_{\text{R}}+\text{P}_{\text{R}}\cdot\text{P}{_\bar{\text{R}}}\cdot\text{P}{_\bar{\text{R}}}$

$=\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}+\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}+\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{4}{6}+\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{2}{6}$

$=\frac{60+60+60+30}{336}=\frac{210}{336}$

$\text{P}(\text{E}_1\cap\text{E}_2)=\text{P}{_\text{R}}\cdot\text{P}{_\bar{\text{R}}}\cdot\text{P}{_\text{R}}+\text{P}{_\text{R}}\cdot\text{P}{_\text{R}}\cdot\text{P}{_\bar{\text{R}}}$

$=\frac{5}{8}\cdot\frac{3}{7}\cdot\frac{4}{6}+\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}=\frac{120}{336}$

$\therefore\text{P}\Big(\frac{\text{E}_2}{\text{E}_1}\Big)=\frac{\text{P}(\text{E}_1\cap\text{E}_2)}{\text{P}(\text{E}_1)}$

$=\frac{\frac{120}{336}}{\frac{210}{336}}=\frac{4}{7}$

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