Question
A bag contains 10 mangoes out of which 4 are rotten, two mangoes are taken out together. If one of them is found to be good, the probability that other also good is-
Answer: Option A
Out of mangoes, 4 mangoes are rotten
∴ Required probability
$$\eqalign{
& = \frac{{{}^6{C_2}}}{{{}^{10}{C_2}}} \cr
& = \frac{{\frac{{6!}}{{2!\left( {6 - 2} \right)!}}}}{{\frac{{10!}}{{2!\left( {10 - 2} \right)!}}}} \cr
& = \frac{{\frac{{6!}}{{2!4!}}}}{{\frac{{10!}}{{2! \times 8!}}}} \cr
& = \frac{{\frac{{6 \times 5}}{{1 \times 2}}}}{{\frac{{10 \times 9}}{{1 \times 2}}}} \cr
& = \frac{{6 \times 5}}{{10 \times 9}} \cr
& = \frac{1}{3} \cr} $$
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Out of mangoes, 4 mangoes are rotten
∴ Required probability
$$\eqalign{
& = \frac{{{}^6{C_2}}}{{{}^{10}{C_2}}} \cr
& = \frac{{\frac{{6!}}{{2!\left( {6 - 2} \right)!}}}}{{\frac{{10!}}{{2!\left( {10 - 2} \right)!}}}} \cr
& = \frac{{\frac{{6!}}{{2!4!}}}}{{\frac{{10!}}{{2! \times 8!}}}} \cr
& = \frac{{\frac{{6 \times 5}}{{1 \times 2}}}}{{\frac{{10 \times 9}}{{1 \times 2}}}} \cr
& = \frac{{6 \times 5}}{{10 \times 9}} \cr
& = \frac{1}{3} \cr} $$
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