A student has to write the words ABILITY,PROB | vedclass

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(A) Let $E_1, E_2, E_3, E_4$ be the events that the boy wrote the words $ABILITY$,$PROBABILITY$,$FACILITY$,and $MOBILITY$ respectively. Since he choos

Vedclass · Accepted Answer

$\frac{21}{116}$

Vedclass · Answer

(A) Let $E_1, E_2, E_3, E_4$ be the events that the boy wrote the words $ABILITY$,$PROBABILITY$,$FACILITY$,and $MOBILITY$ respectively. Since he chooses one word out of four,$P(E_1) = P(E_2) = P(E_3) = P(E_4) = \frac{1}{4}$. Let $A$ be the event that '$LI$' is left after erasing. - For $ABILITY$ ($7$ letters),there are $7-1 = 6$ pairs of consecutive letters. Only one pair is '$LI$'. So,$P(A|E_1) = \frac{1}{6}$. - For $PROBABILITY$ ($11$ letters),there are $11-1 = 10$ pairs of consecutive letters. Only one pair is '$LI$'. So,$P(A|E_2) = \frac{1}{10}$. - For $FACILITY$ ($8$ letters),there are $8-1 = 7$ pairs of consecutive letters. Only one pair is '$LI$'. So,$P(A|E_3) = \frac{1}{7}$. - For $MOBILITY$ ($8$ letters),there are $8-1 = 7$ pairs of consecutive letters. Only one pair is '$LI$'. So,$P(A|E_4) = \frac{1}{7}$. Using Bayes' theorem: $P(E_2|A) = \frac{P(E_2)P(A|E_2)}{\sum_{i=1}^{4} P(E_i)P(A|E_i)} = \frac{\frac{1}{4} 	imes \frac{1}{10}}{\frac{1}{4}(\frac{1}{6} + \frac{1}{10} + \frac{1}{7} + \frac{1}{7})} = \frac{\frac{1}{10}}{\frac{1}{6} + \frac{1}{10} + \frac{2}{7}} = \frac{\frac{1}{10}}{\frac{35+21+60}{210}} = \frac{\frac{1}{10}}{\frac{116}{210}} = \frac{21}{116}$.

$A$ student 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 $PROBABILITY$ is

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