Q78 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics

Astudent has to write the words ABILITY, PROBABILITY, FACILITY, MOBILITY. He wrote one word and erased all the letters in it except two consecutive letters. If ' LI ' is left after erasing then the probability that the boy wrote the word PROBABLLITY is

  • Correct answer
    A
    $\frac{21}{116}$
  • B
    $\frac{72}{116}$
  • C
    $\frac{3}{5}$
  • D
    $\frac{2}{3}$

Solution

Let $\mathrm{E}_1, \mathrm{E}_2, \mathrm{E}_3, \mathrm{E}_4$ be the events that boy wrote the words ABILITY, PROBABILITY, FACILITY, MOBILITY respectively. So, $P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=P\left(E_4\right)=\frac{1}{4}$ A be the event that "LI' left after erasing $\begin{aligned} & \text { So, } P\left(\frac{\mathrm{~A}}{\mathrm{E}_1}\right)=\frac{1}{6}, \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_2}\right)=\frac{1}{10} \\ & \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_3}\right)=\frac{1}{7}, \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_4}\right)=\frac{1}{7}\end{aligned}$ Using, Baye's theorem $\begin{aligned} & P\left(\frac{E_2}{A}\right)=\frac{P\left(E_2\right) P\left(\frac{A}{E_2}\right)}{P\left(E_1\right) P\left(\frac{A}{E_1}\right)+P\left(E_2\right) P\left(\frac{A}{E_2}\right)+P\left(E_3\right) P\left(\frac{A}{E_3}\right)+P\left(E_4\right) P\left(\frac{A}{E_4}\right)}\end{aligned}$
$=\frac{\frac{1}{4} \times \frac{1}{10}}{\frac{1}{4} \times \frac{1}{6}+\frac{1}{4} \times \frac{1}{10}+\frac{1}{4} \times \frac{1}{7}+\frac{1}{4} \times \frac{1}{7}}=\frac{21}{116}$

Hence, the correct answer is option: 'A'.

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