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Q.(Continuing the doctor case study of Q.36) When he arrives, he is late, what is the probability that he comes by train?

Haryana BsehBSEH Intermediate Board 2026Subjective· 2mImportance★★★★★
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Apply Bayes' theorem: P(E1∣L)=P(E1)P(L∣E1)∑P(Ei)P(L∣Ei)P(E_1|L)=\dfrac{P(E_1)P(L|E_1)}{\sum P(E_i)P(L|E_i)}.

From the case study: P(E1)=310P(E_1)=\dfrac{3}{10}, P(E2)=15P(E_2)=\dfrac{1}{5}, P(E3)=110P(E_3)=\dfrac{1}{10}, P(E4)=25P(E_4)=\dfrac{2}{5}; P(L∣E1)=14P(L|E_1)=\dfrac{1}{4}, P(L∣E2)=13P(L|E_2)=\dfrac{1}{3}, P(L∣E3)=112P(L|E_3)=\dfrac{1}{12}, P(L∣E4)=0P(L|E_4)=0.

Numerator: P(E1)P(L∣E1)=310×14=340P(E_1)P(L|E_1)=\dfrac{3}{10}\times\dfrac{1}{4}=\dfrac{3}{40}

Denominator (total probability of being late):

P(E1)P(L∣E1)+P(E2)P(L∣E2)+P(E3)P(L∣E3)+P(E4)P(L∣E4)P(E_1)P(L|E_1)+P(E_2)P(L|E_2)+P(E_3)P(L|E_3)+P(E_4)P(L|E_4)

=340+15×13+110×112+0=340+115+1120=\dfrac{3}{40}+\dfrac{1}{5}\times\dfrac{1}{3}+\dfrac{1}{10}\times\dfrac{1}{12}+0=\dfrac{3}{40}+\dfrac{1}{15}+\dfrac{1}{120}

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