Question

A box contains 10 mangoes out of which 4 are rotten. 2 mangoes are taken out together. If one of them is found to be good, what is the probability that the other is also good?$$$$

Answer 1

We denote the event of selection of a good first mango as A and good second mango as B. We first find the probability of both mangoes being good as $P\left( A\bigcap B \right)$ and then we find the second mango being good $P\left( B \right)$. The required probability is obtained using the formula $P\left( B|A \right)=\dfrac{P\left( A\bigcap B \right)}{P\left( B \right)}$.$$$$

Complete step-by-step answer:
Probability in simple words is the possibility of an event to occur. We know from definition of probability that if there is $n\left( A \right)$ number of ways of event $A$ occurring or favourable outcomes and $n\left( S \right)$ is the size of the sample space or total outcomes then the probability of the event $A$ occurring is $\dfrac{n\left( A \right)}{n\left( S \right)}$. We also know that the probability of en event B happening subjected to A happening is
$P\left( B|A \right)=\dfrac{P\left( A\bigcap B \right)}{P\left( B \right)}$

Let us denote the event of selection of a first mango as A and second mango as B. We note that the box contains 10 mangoes out of which 4 are rotten and hence 6 are good. We are asked probability of the second mango is good subjected to condition the first condition is good which is $P\left( B|A \right)$$$$$

We find that out of the selected mangoes both are good. We can choose 2 mangos out of 10 mangoes in $n\left( A\bigcap B \right){{=}^{6}}{{C}_{2}}$ ways. The probability of selecting two good mangos at the same time is,
$P\left( A\bigcap B \right)=\dfrac{^{6}{{C}_{2}}}{^{10}{{C}_{2}}}=\dfrac{1}{3}$

We need to find the probability of B which is the selection of only the second mango as good mango. We see that we one that we select a good mango after the first one is rotten or good. So the sum of such cases are $n\left( B \right){{=}^{6}}{{C}_{1}}{{\times }^{4}}{{C}_{2}}{{+}^{6}}{{C}_{2}}$. So the probability of second mango being good is
$P\left( B \right)=\dfrac{n\left( B \right)}{n\left( S \right)}=\dfrac{^{6}{{C}_{1}}{{\times }^{4}}{{C}_{2}}{{+}^{6}}{{C}_{2}}}{^{10}{{C}_{2}}}=\dfrac{39}{45}$

So the required probability is
$P\left( B|A \right)=\dfrac{P\left( A\bigcap B \right)}{P\left( B \right)}=\dfrac{1}{3}\times \dfrac{45}{39}=\dfrac{5}{13}$

Note: We can alternatively find the probability by listing all the outcomes as pairs. If the events are A and B are independent which means the event A does not influence the other event B then we have $P\left( A\bigcap B \right)=P\left( A \right)P\left( B \right)$ since $P\left( B|A \right)=P\left( B \right)$ .