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Probability Question

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gmatian
  • Authority 25
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gmatian said:

Hi,

can nybdy give me the solution for this?

In a group of 20 office buildings, 5 have been inspected this year. Of the uninspected buildings, 3 have fire code violations. If the city randomly inspects two uninspected buildings, what is the probability that at least one of them has fire code violations?

Answer is supposed to be 13/45…

plz help…

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  • Posted 4 months ago.
gauravjain26
  • Authority 592
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gauravjain26 said:

Solution:

Total number of ways in which two buildings out of 15 buildings(20- 5) can be selected = 15 X 14 (i.e. 15 p 2)

probability of getting 0 building selected from code violating ones= \frac{12 * 11}{15 * 14}

Probability of getting atleast one from violating codes = 1 – (probability of getting 0 buildings from code violation)

=> 1 - \frac{(12 X 11)}{(15 * 14)}

=> \frac{13}{35}


for future queries please look for the solution in CAT or Gmat or SAT community.

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  • Posted 2 months ago.
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