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Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability1 Mark
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
    $\frac{1}{3}$
  • ✓
    $\frac{4}{7}$
  • C
    $\frac{15}{28}$
  • D
    $\frac{5}{28}$
✓

Answer

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

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