Classical probability definition

We select two numbers, $$a$$ and then $$b$$ from the first 100 natural numbers without replacement.
Thus the total number of ordered selections is $$100\times\ 99=9900$$

We need $$P\left(a-b\ge\ 10\right)$$

$$\therefore$$ $$\left(a\ge b+\ 10\right)$$

So for every chosen $$a$$, the value of $$b$$ must be at least 10 or less than $$a$$.

The smallest possible $$a$$ satisfying the condition is 11.

For each $$a$$, the possible values of $$b$$ are:

1,2,3,…,a−10

Number of choices for $$b$$: $$a-10$$

Favourable outcomes:

$$\Sigma_{a=11}^{a=100}\left(a-10\right)$$

Let k=a-10

$$\therefore$$ $$\Sigma_{k=1}^{k=90}\left(k\right)$$

$$sum=\dfrac{\left(90\left(90+1\right)\right)}{2}=4095$$

$$P=\dfrac{4095}{9900}$$

When we simplify for coprime, we get $$\dfrac{4095}{9900}=\dfrac{91}{220}$$

$$\therefore$$ m=91 and n=220 

$$m+n=91+220=311$$

Hence, our answer is 311.

Let us determine the total counts of the balls and find the probabilities for each event.

Box I contains:

$$ \text{Red balls} = 6 $$

$$ \text{Green balls} = 9 $$

$$ \text{Total balls in Box I} = 6 + 9 = 15 $$

Box II contains:

$$ \text{Red balls} = 8 $$

$$ \text{Green balls} = 12 $$

$$ \text{Total balls in Box II} = 8 + 12 = 20 $$

All balls are mixed together:

$$ \text{Total Red balls } (F_1) = 6 + 8 = 14 $$

$$ \text{Total Green balls } (F_2) = 9 + 12 = 21 $$

$$ \text{Overall Total balls} = 15 + 20 = 35 $$

Step 1: Calculate Individual Probabilities

The probability that a chosen ball belongs to Box I is:

$$ P(E_1) = \frac{15}{35} = \frac{3}{7} $$

The probability that a chosen ball belongs to Box II is:

$$ P(E_2) = \frac{20}{35} = \frac{4}{7} $$

The probability that a chosen ball is red is:

$$ P(F_1) = \frac{14}{35} = \frac{2}{5} $$

The probability that a chosen ball is green is:

$$ P(F_2) = \frac{21}{35} = \frac{3}{5} $$

Step 2: Evaluate Independence of $$ E_1 $$ and $$ F_1 $$

The intersection probability $$ P(E_1 \cap F_1) $$ represents choosing a red ball that came from Box I:

$$ P(E_1 \cap F_1) = \frac{6}{35} $$

Now, let us calculate the product of their individual probabilities:

$$ P(E_1) \cdot P(F_1) = \frac{3}{7} \cdot \frac{2}{5} = \frac{6}{35} $$

Since $$ P(E_1 \cap F_1) = P(E_1) \cdot P(F_1) $$, the events $$ E_1 $$ and $$ F_1 $$ are independent. Therefore, Statement A is TRUE.

Step 3: Evaluate Independence of $$ E_2 $$ and $$ F_2 $$

The intersection probability $$ P(E_2 \cap F_2) $$ represents choosing a green ball that came from Box II:

$$ P(E_2 \cap F_2) = \frac{12}{35} $$

Now, let us calculate the product of their individual probabilities:

$$ P(E_2) \cdot P(F_2) = \frac{4}{7} \cdot \frac{3}{5} = \frac{12}{35} $$

Since $$ P(E_2 \cap F_2) = P(E_2) \cdot P(F_2) $$, the events $$ E_2 $$ and $$ F_2 $$ are independent. This means any statement claiming they are dependent is FALSE.

Step 4: Evaluate Conditional Probabilities

Let us calculate $$ P(F_1 \mid E_1) $$:

$$ P(F_1 \mid E_1) = \frac{P(E_1 \cap F_1)}{P(E_1)} = \frac{\frac{6}{35}}{\frac{15}{35}} = \frac{6}{15} = \frac{2}{5} $$

Let us calculate $$ P(F_1 \mid E_2) $$:

$$ P(F_1 \mid E_2) = \frac{P(E_2 \cap F_1)}{P(E_2)} = \frac{\frac{8}{35}}{\frac{20}{35}} = \frac{8}{20} = \frac{2}{5} $$

Since $$ P(F_1 \mid E_1) = P(F_1 \mid E_2) = \frac{2}{5} $$, the conditional probabilities are equal. Therefore, Statement C is FALSE.

Let us check the last comparison involving $$ P(F_2 \mid E_2) $$:

$$ P(F_2 \mid E_2) = \frac{P(E_2 \cap F_2)}{P(E_2)} = \frac{\frac{12}{35}}{\frac{20}{35}} = \frac{12}{20} = \frac{3}{5} $$

Comparing the values, $$ \frac{2}{5} $$ is not greater than $$ \frac{3}{5} $$, making any statement asserting $$ P(F_1 \mid E_1) > P(F_2 \mid E_2) $$ FALSE.

Conclusion

The correct statements are:

A. The events $$ E_1 $$ and $$ F_1 $$ are independent.

Suppose Mr. A starts on channel $$C.$$

At each minute, he must switch to a different channel.

Hence from any channel, there are always

$$4$$

possible choices.

We need the probability that he comes back to channel $$C$$ for the first time exactly at the end of the fifth minute.

That means:

- During the first $$4$$ minutes, he should never return to $$C$$.
- On the fifth minute, he must return to $$C$$.

Total number of possible paths in $$5$$ minutes:

$$4^5$$

Now count favourable paths.

Minute 1:

He must leave $$C.$$

Number of choices:

$$4$$

Minute 2:

He should not return to $$C.$$

Also, he cannot stay on the same channel.

Out of the $$4$$ available channels, one is $$C$$ which is forbidden.

Hence choices:

$$3$$

Similarly,

Minute 3:

$$3$$

choices.

Minute 4:

$$3$$

choices.

Minute 5:

He must return to $$C.$$

Exactly

$$1$$

choice.

Therefore, favourable paths:

$$4\times3\times3\times3\times1$$

$$=108$$

Hence required probability:

$$\frac{108}{4^5}$$

$$=\frac{108}{1024}$$

$$=\frac{27}{256}$$

Thus,

$$m=27,\quad n=256$$

Therefore,

$$m+n=27+256$$

$$=283$$

Hence, the required answer is

$$\boxed{283}$$