ISI2019-MMA-10

Question

The chance of a student getting admitted to colleges $A$ and $B$ are $60\%$ and $40\%$, respectively. Assume that the colleges admit students independently. If the student is told that he has been admitted to at least one of these colleges, what is the probability that he has got admitted to college $A$?

• A. $3/5$
• B. $5/7$
• C. $10/13$
• D. $15/19$

Comments

D is correct. Even I arrived at 3/5 which is wrong

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yes..Bayes theorem cannot apply here, because of atleastword. right??

Answer 1

P(getting admitted to A $\cup$ getting admitted to B)=P(getting admitted to A)+P(getting admitted to B)-P(getting admitted to A $\cap$ getting admitted to B)=$1-\frac{3}{5} \times \frac{2}{5} =\frac{19}{25}$ . This quantity has to deducted as it is being double counted in the sample space.

Therefore, P(getting admitted to A given he is admitted to atleast one of the colleges)=$\frac{\frac{3}{5}}{\frac{19}{25}}=\frac{3}{5} \times \frac{25}{19}=\frac{15}{19}$