Question

An urn contains 3 white, 4 red and 5 black balls. Two balls are drawn one by one without replacement. What is the probability that at least one ball is black?

Answer 1

Hint: Find the number of non-black balls present. Thus to get the probability of at least one black ball, find 1 minus the number of non – black balls.

Complete step-by-step answer:
Given the total number of white balls = 3.
Total number of Red balls = 4.
The number of black balls in the urn = 5.
Thus the total number of balls = number of white balls + number of red balls + number of black balls.
                                                       = 3 + 4 + 5 = 12
Thus total balls in the urn = 12.
Out of the total 12 balls, the number of balls which are not black in color
= Total balls – number of black balls
= 12 – 5
=7
$$\therefore$$ Number of non –black balls = 7.
Now we have to draw out 2 balls without replacement.
P (at least one black ball) = 1 – P (none is black) –(1)
First let us calculate the probability that none of the 2 balls are black i.e. both the balls are of white and red.
The first withdrawal of black balls = number of non – black balls/ total balls = $${\displaystyle \frac{7}{12}}$$.
Now let the second withdrawal of the black ball, if a black ball has already been retired (without placement).
= number of non-black – 1/ total - 1$$={\displaystyle \frac{7-1}{12-1}}={\displaystyle \frac{6}{11}}$$
Now let us calculate the probability of at least 1 ball is black.
$$\therefore$$ P (at least one black ball) = 1 – P (none is black)
                                                   $$\begin{array}{rl}& =1-({\displaystyle \frac{7}{12}}\times {\displaystyle \frac{6}{11}})\\ & =1-\left({\displaystyle \frac{7\times 6}{12\times 11}}\right)=1-{\displaystyle \frac{42}{132}}={\displaystyle \frac{132-42}{132}}\\ & ={\displaystyle \frac{15}{22}}\end{array}$$
Thus we got the probability of getting at least one black ball $$={\displaystyle \frac{15}{22}}$$ .
Note: We can also solve this question by simple combination.
P (at least one black ball) = 1 – (no black ball)
                                              $$\begin{array}{rl}& =1-{\displaystyle \frac{{}^7{C}_2}{{}^{12}{C}_2}}\\ & =1-{\displaystyle \frac{{\displaystyle \frac{7!}{(7-2)!2!}}}{{\displaystyle \frac{12!}{(12-2)!2!}}}}=1-{\displaystyle \frac{{\displaystyle \frac{7!}{5!2!}}}{{\displaystyle \frac{12!}{10!2!}}}}\\ & =1-{\displaystyle \frac{7\times {\displaystyle \frac{6}{2}}}{12\times {\displaystyle \frac{11}{2}}}}=1-{\displaystyle \frac{7\times 3}{11\times 6}}\\ & =1-{\displaystyle \frac{21}{66}}\end{array}$$
$$\therefore$$ P (at least one black ball) $$={\displaystyle \frac{66-21}{66}}={\displaystyle \frac{15}{22}}$$.