Question

In a bag there are six balls of unknown colors, three balls are drawn and found to be black. Find the chance that no black ball is left in the bag.
A. $\dfrac{1}{{35}}$
B. $\dfrac{1}{3}$
C. $\dfrac{2}{{35}}$
D. $\dfrac{1}{5}$

Answer 1

Here we will take all the possibilities of black balls in the bag where we will use the information that there can be more than three black balls in the bag. We find all the possible ways in which we can take out three black balls( from six black, from five black, from four black and from three black balls in the bag) and then use the method of probability to find the probability of three black balls only in the bag because those three black balls are drawn.

Complete step-by-step answer:
Total number of balls in the bag $= 6$
Since, we know three black balls are drawn from the bag
Therefore we find outcomes where we draw $3$ black balls from $6,5,4,3$ black balls in the bag separately.
Number of ways three balls are chosen is given by the number of black balls from which we can choose/ total number of balls.
As we fill the requirement for the first ball we reduce the total number of balls by one and reduce the number of black balls in each case by one, similarly we do so after filling the requirement for the second ball.
First case:
Probability of drawing $3$ black balls from $6$ black balls $= \dfrac{6}{6} \times \dfrac{5}{5} \times \dfrac{4}{4} = 1$
Second case:
Probability of drawing $3$ black balls from $5$ black balls $= \dfrac{5}{6} \times \dfrac{4}{5} \times \dfrac{3}{4} = \dfrac{1}{2}$
Third case:
Probability of drawing $3$ black balls from $4$ black balls $= \dfrac{4}{6} \times \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{1}{5}$
Fourth case:
Probability of drawing $3$black balls from $3$ black balls $= \dfrac{3}{6} \times \dfrac{2}{5} \times \dfrac{1}{4} = \dfrac{1}{{20}}$
Therefore now we sum up all the possibilities which give us total number of ways in which three black balls can be drawn from three black, four black, five black and six black balls in the bag
Total number of ways $= 1 + \dfrac{1}{2} + \dfrac{1}{5} + \dfrac{1}{{20}}$
Taking the LCM $= \dfrac{{20 + 10 + 4 + 1}}{{20}} = \dfrac{{35}}{{20}}$
Now we find the probability of no black ball in the bag which is the probability of three black balls only from the fourth case.
Probability $=$ number of ways to choose three black balls from three black balls in the bag / total number of ways to choose three black balls from the bag
Probability $= \dfrac{{\left( {\dfrac{1}{{20}}} \right)}}{{\left( {\dfrac{{35}}{{20}}} \right)}}$
$$
= \dfrac{1}{{20}} \times \dfrac{{20}}{{35}} \\
= \dfrac{1}{{35}} \\

Thus, option A is correct. **Note:** Alternate method: We can also find the number of ways to choose three black balls using the method of combination Combination gives us the formula $$^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$$ We know number of total ways to choose three black balls from total of six black balls is $$^6{C_3}$$ And number of ways to choose three black balls from $$n$$black balls will be given by $$^n{C_3}$$, $$n = 3,4,5,6$$ First case: Probability of drawing $$3$$ black balls from $$6$$black balls $$ = \dfrac{{^6{C_3}}}{{^6{C_3}}} = 1$$ Second case: Probability of drawing $$3$$ black balls from $$5$$ black balls $$ = \dfrac{{^5{C_3}}}{{^6{C_3}}}$$ $$ = \dfrac{{\dfrac{{5!}}{{(5 - 3)!3!}}}}{{\dfrac{{6!}}{{(6 - 3)!3!}}}} \\\ = \dfrac{{\dfrac{{5 \times 4 \times 3!}}{{2! \times 3!}}}}{{\dfrac{{6 \times 5 \times 4 \times 3!}}{{3! \times 3!}}}} \\\ = \dfrac{{5 \times 4 \times 3!}}{{2! \times 3!}} \times \dfrac{{3! \times 3!}}{{6 \times 5 \times 4 \times 3!}} \\\

Cancel terms from numerator and denominator and use $n! = n \times (n - 1)!$ to open the factorial.
$$
= \dfrac{{5 \times 4 \times 3 \times 2!}}{{6 \times 5 \times 4 \times 2!}} \\
= \dfrac{{5 \times 4 \times 3}}{{6 \times 5 \times 4}} \\
= \dfrac{1}{2} \\

Third case: Probability of drawing $$3$$ black balls from $$4$$ black balls $$ = \dfrac{{^4{C_3}}}{{^6{C_3}}}$$ $$ = \dfrac{{\dfrac{{4!}}{{(4 - 3)!3!}}}}{{\dfrac{{6!}}{{(6 - 3)!3!}}}} \\\ = \dfrac{{\dfrac{{4 \times 3!}}{{1! \times 3!}}}}{{\dfrac{{6 \times 5 \times 4 \times 3!}}{{3! \times 3!}}}} \\\ = \dfrac{{4 \times 3!}}{{1! \times 3!}} \times \dfrac{{3! \times 3!}}{{6 \times 5 \times 4 \times 3!}} \\\

Cancel terms from numerator and denominator and use $n! = n \times (n - 1)!$ to open the factorial.
$$
= \dfrac{{4 \times 3 \times 2}}{{6 \times 5 \times 4}} \\
= \dfrac{1}{5} \\

Fourth case: Probability of drawing $$3$$ black balls from $$3$$ black balls $$ = \dfrac{{^3{C_3}}}{{^6{C_3}}}$$ $$ = \dfrac{{\dfrac{{3!}}{{(3 - 3)!3!}}}}{{\dfrac{{6!}}{{(6 - 3)!3!}}}} \\\ = \dfrac{{\dfrac{{3!}}{{0! \times 3!}}}}{{\dfrac{{6 \times 5 \times 4 \times 3!}}{{3! \times 3!}}}} \\\ = \dfrac{{3!}}{{0! \times 3!}} \times \dfrac{{3! \times 3!}}{{6 \times 5 \times 4 \times 3!}} \\\

Cancel terms from numerator and denominator and use $n! = n \times (n - 1)!$ to open the factorial. Also $0! = 1$
$$
= \dfrac{{3 \times 2}}{{6 \times 5 \times 4}} \\
= \dfrac{1}{{20}} \\

Now probability if no black ball in the bag is $$ = \dfrac{{\dfrac{{^3{C_3}}}{{^6{C_3}}}}}{{\dfrac{{^3{C_3}}}{{^6{C_3}}} + \dfrac{{^3{C_3}}}{{^6{C_3}}} + \dfrac{{^3{C_3}}}{{^6{C_3}}} + \dfrac{{^3{C_3}}}{{^6{C_3}}}}}$$ Substituting the values we get Probability $$ = \dfrac{{\dfrac{1}{{20}}}}{{1 + \dfrac{1}{2} + \dfrac{1}{5} + \dfrac{1}{{20}}}} = \dfrac{{\dfrac{1}{{20}}}}{{\dfrac{{20 + 10 + 4 + 1}}{{20}}}} = \dfrac{{\dfrac{1}{{20}}}}{{\dfrac{{35}}{{20}}}} = \dfrac{1}{{20}} \times \dfrac{{20}}{{35}} = \dfrac{1}{{35}}$$ Thus, Option A is correct.