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**Hint:**According to the given in the question that a bag contains 6 red marbles and 4 blue marbles. A marble is drawn at random and not replaced. Two further drawings are made, again without replacement. So, first of all to determine the probability of drawing three red marbles so for that we have to choose the one red marble from the given red marbles and the we have to find the one red marble from all the given ten red marbles and as given two more red marbles are drawn and repletion is not allowed hence, now we have to choose the second marbles from the remaining marbles except one red marble which is already drawn and we will repeat the same process for the remaining third red marble. We can choose the marble from the given marbles with the help of the formula as given below:

**Formula used:**$ \Rightarrow C_r^n = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}..................(A)$

Where n is the number of marbles and r is the number of required marbles for which we want to find the probability.

Now, to find the probability that three blue marbles are drawn so, we have to choose the one blue marble from the given blue marbles and the we have to find the one blue marble from all the given ten blue marbles and as given two more blue marbles are drawn and repletion is not allowed hence, now we have to choose the second marbles from the remaining marbles except one blue marble which is already drawn and we will repeat the same process for the remaining third blue marble.

Now, to find that no red marble is drawn we have to find the probability that all the blue marbles are chosen,

Now, to find at least one red marble we have to subtract the obtained probability that all the blue marbles are chosen by one.

**Complete step-by-step solution:**

Step 1: First of all we have to determine the probability that one red marble is drawn from the bag hence, with the help of the formula (A),

$ \Rightarrow \dfrac{{C_1^6}}{{C_1^{10}}} = \dfrac{{\dfrac{{6!}}{{1!\left( {6 - 1} \right)!}}}}{{\dfrac{{10!}}{{1!\left( {10 - 1} \right)!}}}}$

On solving the expression as obtained just above,

$

\Rightarrow \dfrac{{C_1^6}}{{C_1^{10}}} = \dfrac{{\dfrac{{6 \times 5!}}{{1!\left( 5 \right)!}}}}{{\dfrac{{10 \times 9!}}{{1!\left( 9 \right)!}}}} \\

\Rightarrow \dfrac{{C_1^6}}{{C_1^{10}}} = \dfrac{6}{{10}}...........(1)

$

Now, same as we have to determine the probability for the second red marble which is drawn from the bag as mentioned in the solution hint.

$ \Rightarrow \dfrac{{C_1^5}}{{C_1^9}} = \dfrac{{\dfrac{{5!}}{{1!\left( {5 - 1} \right)!}}}}{{\dfrac{{9!}}{{1!\left( {9 - 1} \right)!}}}}$

On solving the expression as obtained just above,

$

\Rightarrow \dfrac{{C_1^5}}{{C_1^9}} = \dfrac{{\dfrac{{5 \times 4!}}{{1!\left( 4 \right)!}}}}{{\dfrac{{9 \times 8!}}{{1!\left( 8 \right)!}}}} \\

\Rightarrow \dfrac{{C_1^6}}{{C_1^{10}}} = \dfrac{5}{9}...........(2)

$

Now, same as we have to determine the probability for the third red marble which is drawn from the bag as mentioned in the solution hint.

$ \Rightarrow \dfrac{{C_1^4}}{{C_1^8}} = \dfrac{{\dfrac{{4!}}{{1!\left( {4 - 1} \right)!}}}}{{\dfrac{{8!}}{{1!\left( {8 - 1} \right)!}}}}$

On solving the expression as obtained just above,

$

\Rightarrow \dfrac{{C_1^4}}{{C_1^8}} = \dfrac{{\dfrac{{4 \times 3!}}{{1!\left( 3 \right)!}}}}{{\dfrac{{8 \times 7!}}{{1!\left( 7 \right)!}}}} \\

\Rightarrow \dfrac{{C_1^6}}{{C_1^{10}}} = \dfrac{4}{8}...........(3)

$

Hence, probability that all three red marbles are drawn at random without repetition we have to multiply all (1), (2), and (3).

$

\Rightarrow \dfrac{6}{{10}} \times \dfrac{5}{9} \times \dfrac{4}{8} = \dfrac{{120}}{{720}} \\

= \dfrac{1}{6}

$

Step 2: First of all we have to determine the probability that one blue marble is drawn from the bag hence, with the help of the formula (A),

$ \Rightarrow \dfrac{{C_1^4}}{{C_1^{10}}} = \dfrac{{\dfrac{{4!}}{{1!\left( {4 - 1} \right)!}}}}{{\dfrac{{10!}}{{1!\left( {10 - 1} \right)!}}}}$

On solving the expression as obtained just above,

$

\Rightarrow \dfrac{{C_1^4}}{{C_1^{10}}} = \dfrac{{\dfrac{{4 \times 3!}}{{1!\left( 3 \right)!}}}}{{\dfrac{{10 \times 9!}}{{1!\left( 9 \right)!}}}} \\

\Rightarrow \dfrac{{C_1^4}}{{C_1^{10}}} = \dfrac{4}{{10}}...........(4)

$

Now, same as we have to determine the probability for the second blue marble which is drawn from the bag as mentioned in the solution hint.

$ \Rightarrow \dfrac{{C_1^3}}{{C_1^9}} = \dfrac{{\dfrac{{3!}}{{1!\left( {3 - 1} \right)!}}}}{{\dfrac{{9!}}{{1!\left( {9 - 1} \right)!}}}}$

On solving the expression as obtained just above,

$

\Rightarrow \dfrac{{C_1^3}}{{C_1^9}} = \dfrac{{\dfrac{{3 \times 2!}}{{1!\left( 2 \right)!}}}}{{\dfrac{{9 \times 8!}}{{1!\left( 8 \right)!}}}} \\

\Rightarrow \dfrac{{C_1^3}}{{C_1^{10}}} = \dfrac{3}{9}...........(5)

$

Now, same as we have to determine the probability for the third blue marble which is drawn from the bag as mentioned in the solution hint.

$ \Rightarrow \dfrac{{C_1^2}}{{C_1^8}} = \dfrac{{\dfrac{{2!}}{{1!\left({2-1} \right)!}}}}{{\dfrac{{8!}}{{1!\left( {8 - 1} \right)!}}}}$

On solving the expression as obtained just above,

$

\Rightarrow \dfrac{{C_1^2}}{{C_1^8}} = \dfrac{{\dfrac{{2 \times 1!}}{{1!\left( 1 \right)!}}}}{{\dfrac{{8 \times 7!}}{{1!\left( 7 \right)!}}}} \\

\Rightarrow \dfrac{{C_1^2}}{{C_1^{10}}} = \dfrac{2}{8}...........(6)

$

Hence, probability that all three blue marbles are drawn at random without repetition we have to multiply all (4), (5), and (6).

$

= \dfrac{4}{{10}} \times \dfrac{3}{9} \times \dfrac{2}{8} \\

= \dfrac{{24}}{{720}} \\

= \dfrac{1}{3}

$

Step 3: Now, to find the probability that no red ball is drawn from the bad then we have to multiply the expressions (4), (5), and (6) as explained in the solution hint. Hence,

Probability that no red marble is drawn is

$

= \dfrac{4}{{10}} \times \dfrac{3}{9} \times \dfrac{2}{8} \\

= \dfrac{{24}}{{720}} \\

= \dfrac{1}{3}

$

Step 4: Now, to find that at least one red marble we have to subtract obtained probability that all the blue marbles are chosen by one as mentioned in the solution hint, Hence, with the help of step 3.

$

= 1 - \dfrac{1}{3} \\

= \dfrac{2}{3}

$

**Hence, with the help of formula (A) we have obtained:**

A) Three red marbles = $\dfrac{1}{6}$

B) Three blue marbles = $\dfrac{1}{3}$

C) No red marbles = $\dfrac{1}{6}$

D) At least one red marble = $\dfrac{2}{3}$

A) Three red marbles = $\dfrac{1}{6}$

B) Three blue marbles = $\dfrac{1}{3}$

C) No red marbles = $\dfrac{1}{6}$

D) At least one red marble = $\dfrac{2}{3}$

**Note:**If the probability of an event to occur is p then the probability an event not to occur can be obtained by subtracting the probability of an event to occur which is p by 1.

If repetition is not allowed that means if one marble is drawn from the bag then for the next marble we can’t consider the precious marble which is already drawn.

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