# Easy Probability Solved QuestionAptitude Discussion

 Q. Out of 17 applicants 8 boys and 9 girls. Two persons are to be selected for the job. Find the probability that at least one of the selected persons will be a girl.
 ✔ A. 27/34 ✖ B. 25/34 ✖ C. 19/34 ✖ D. 21/34

Solution:
Option(A) is correct

The events of selection of two person is redefined as (i) first is a girl AND second is a boy OR (ii) first is boy AND second

is a girl OR (iii) first is a girl and second is a girl.

So the required probability:

$=\left(\dfrac{9}{17}+\dfrac{8}{16}\right)+\left(\dfrac{8}{17}+\dfrac{9}{16}\right)+\left(\dfrac{9}{17}+\dfrac{8}{16}\right)$

$=\dfrac{9}{34}+\dfrac{9}{34}+\dfrac{9}{34}$

$=\dfrac{27}{34}$

Edit: As pointed by Brijesh, final answer has been changed from option (B) to option (A).

Edit 2: For a quick alternative solution, check comment by SHIVAM.

Edit 3: For yet another alternative method using negation approach, check comment by Shrinivas.

## (10) Comment(s)

Anonymous
()

the required probability mentioned on the solution is wrong ...

=(9/17+8/16)+(8/17+9/16)+(9/17+8/16)

for me (9/17+8/16) is not equal to 9/34

Shrinivas
()

ANS = 1- P(No girl is selected)

P(No girl is selected) $= {^8C_2}{ ^{17}C_2}$

$=\dfrac{7}{34}$

Therefore,

ANS $= 1- \dfrac{7}{34}$

$= \dfrac{27}{34}$

SHIVAM
()

You can also do that,

$\dfrac{{^8C_1} \times {^9C_1}}{{^{17}C_2}} + \dfrac{{^8C_0} \times {^9C_2}}{{^{17}C_2}} = \dfrac{27}{34}$

Brijesh
()

$\dfrac{9}{34} + \dfrac{9}{34} + \dfrac{9}{34} =\dfrac{25}{34}$? is it?

Isn't it supposed to be $\dfrac{27}{34}$.

Buggi
()

kya yar itna v mistake mat karo.... ans - 27/34

Vaibhav
()

This is combination, not arrangement

either 1 girl and 1 boy or both being a girl

Priyanka
()

the problem is regarding only selection of applicants not about arrangement. so the ans should be 27/34

Deepika Jain
()

Ravish
()

ans is $27/34$ cz question says 9 girls and solution says 8 girls ..

Bis
()

The arrangement is wrong. I think the answer should be $27/34$. Please recheck.