How to find a possibility of match in Naive Bayes Classifier?
Let $A$ and $O$ be the events apple and orange (fruit). Let $R$ and $S$ be the events red and sphere.

By Bayes' law,
$$\Pr[A | R \wedge S] = \frac{ \Pr[R \wedge S | A] \Pr[A] }{ \Pr[R \wedge S] };$$
$$\Pr[O | R \wedge S] = \frac{ \Pr[R \wedge S | O] \Pr[O] }{ \Pr[R \wedge S] }.$$
By total probability adding to 1, we also have
$$\Pr[A | R \wedge S] + \Pr[O | R \wedge S] = 1$$
and therefore
$$\Pr[R \wedge S | A] \Pr[A] + \Pr[R \wedge S | O] \Pr[O] = \Pr[R \wedge S]$$
so
$$\Pr[A | R \wedge S] = \frac{\Pr[R \wedge S | A] \Pr[A] }{ \Pr[R \wedge S | A] \Pr[A] + \Pr[R \wedge S | O] \Pr[O]}.$$
By the Naive Bayes assumption of independence of features conditioned on class,
$$\Pr[R \wedge S | A] = \Pr[R | A] \Pr[S | A];$$
$$\Pr[R \wedge S | O] = \Pr[R | O] \Pr[S | O].$$
Putting it together, we get
$$\Pr[A | R \wedge S] = \frac{\Pr[R | A] \Pr[S | A] \Pr[A] }{ \Pr[R | A] \Pr[S | A] \Pr[A] + \Pr[R | O] \Pr[S | O] \Pr[O]}.$$

Now plug and chug. From the data $\Pr[R | A] = \frac{ \Pr[R \wedge A] }{ \Pr[A] } = \frac{2}{5}$, etc. Note: I am assuming the empirical distribution is the true distribution, an assumption your teacher probably expects you to make but not made in practice.