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A TextBook Probability Problem:: there're $40$ students in a class among whom $20$ students play football, $25$ students play cricket and $10$ play both. if one is selected who plays football, then determine probability of his playing cricket.

so first i solved it in that way:: among $20$ students who play football, there're $10$ who play cricket also... then the answer is $\frac{10}{20}$ = $\frac{1}{2}$

then i found another solution:: there're total $35$ students who play at least one games, so probability of his playing football is $\frac{20}{35}$ and probability of his playing cricket is $\frac{25}{35}$ , then the answer is $\frac{20}{35}\cdot\frac{25}{35}$ = $\frac{20}{49}$

which one is wrong and why?

The first one is right.
The second one is not right because you can't cut the five students out from the list and you must count them while calculating the probability.
There is a strong evidence for the first answer,I don't know,you know about the probability under a condition or not.For your kind information,I'm showing you that,
\[P\frac {(Cricket)}{(Football)} = \frac {P(Cricket \bigcap Football)}{P(Football)}\]
\[= \frac {\frac {10}{40}}{\frac {20}{40}}\]
\[= \frac {1}{2}\]


here is the probability of playing cricket with respect to playing football.
Do you have any confusion?????
If you have,please reply.

Why do you considering the probability of playing football?
You are already given with his football playing.

kamrul2010 wrote:Why do you considering the probability of playing football?
You are already given with his football playing.

I've considered the probability of playing football because there have been told to consider it.

if one is selected who plays football, then determine probability of his playing cricket.

This is the probability under certain condition.Here playing football is the condition and that's why you must have to consider it.

Oops! :-S

Didn't get it!

whatever,thanks for clearing!