JEE Mains · Maths · STD 12 - 13. probability
A fair die is tossed repeatedly until a six is obtained. Let \(\mathrm{X}\) denote the number of tosses required and let \(\mathrm{a}=\mathrm{P}(\mathrm{X}=3), \mathrm{b}=\mathrm{P}(\mathrm{X} \geq 3)\) and \(\mathrm{c}=\) \(\mathrm{P}(\mathrm{X} \geq 6 \mid \mathrm{X}>3)\). Then \(\frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}\) is equal to
- A \(19\)
- B 12
- C \(14\)
- D \(16\)
Answer & Solution
Correct Answer
(B) 12
Step-by-step Solution
Detailed explanation
\( a=P(X=3)=\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}=\frac{25}{216} \) \( b=P(X \geq 3)=\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^3 \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^4 \cdot \frac{1}{6}+\ldots \ldots \)…
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