Calculate lottery's second prize using combination - lottery probability question.

Your logic for case number (3) does not seem correct.

First, you have to draw 6 balls with exactly 5 balls matching 5 numbers from your set of 6 picked numbers. In total There are C(42,6) ways to draw six balls. How many winning combinations do we have? From your set of 6 balls you can pick five matching balls in C(6,5) ways. The sixth number can be any number from the remaining 42-5=37, minus one (your sixth pick). So the total number of winning combinations is $C(6,5)\times36$. So the probability is:

$$p_1=\frac{36\times{6\choose 5}}{42\choose 6}$$

But you have to draw the seventh number as well from the remaining 42-6=36 numbers without hitting your last (sixth) number. That probability is:

$$p_2=\frac{35}{36}$$

The total proability is:

$$p=p_1\times p_2=\frac{36\times{6\choose 5}}{42\choose 6}\times\frac{35}{36}=\frac{35\times{6\choose 5}}{42\choose 6}=\frac{15}{374699}$$

You can use a similar logic for case (2).

The probability $p_1$ is the same. For the seventh ball we have 36 choices and there is 1 winning ball between them. So the probability $p_2$ is:

$$p_2=\frac{1}{36}$$

...and the final probability for the second prize is:

$$p=p_1\times p_2=\frac{36\times{6\choose 5}}{42\choose 6}\times\frac{1}{36}=\frac{{6\choose 5}}{42\choose 6}=\frac{3}{2622893}$$

(35 times smaller than the probability for the third prize)

For $3^{rd}$ prize, you need to select which 5 numbered balls you will be getting, out of 6 in $^6C_5$ ways (get balls matching to those 5 numbers from 42); and then, select the $6^{th}$ ad $7^{th}$ ball in $^{36}C_2$ ways
(42-6=36, balls not matching any of the 6 numbers)

So, probability of winning $3^{rd}$ prize is $$\frac{^6C_5\cdot^{36}C_2}{^{42}C_7}$$

For $2^{nd}$ prize,
Among first 6, only 5 match, selected their corresponding numbers in $^6C_5$ ways.
First 5 balls drawn match these 5 numbers.
Now, $6^{th}$ ball is selected from 36 (42-6=36, it does not match any of 6 numbers); and number corresponding to last ball to be drawn is already known(one left from 6)

So, probability of winning $2^{nd}$ prize is: $$\frac{^6C_5\cdot^{36}C_1\cdot^{1}C_1}{^{42}C_7}$$

Similarly, For $1^{st}$ prize, $$\frac{^6C_6\cdot^{36}C_1}{^{42}C_7}$$