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 = \[\dfrac{7}{12}\].
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\[=\dfrac{7-1}{12-1}=\dfrac{6}{11}\]
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)
                                                   \[\begin{align}
  & =1-\left( \dfrac{7}{12}\times \dfrac{6}{11} \right) \\
 & =1-\left( \dfrac{7\times 6}{12\times 11} \right)=1-\dfrac{42}{132}=\dfrac{132-42}{132} \\
 & =\dfrac{15}{22} \\
\end{align}\]
Thus we got the probability of getting at least one black ball \[=\dfrac{15}{22}\] .
Note: We can also solve this question by simple combination.
P (at least one black ball) = 1 – (no black ball)
                                              \[\begin{align}
  & =1-\dfrac{{}^{7}{{C}_{2}}}{{}^{12}{{C}_{2}}} \\
 & =1-\dfrac{\dfrac{7!}{\left( 7-2 \right)!2!}}{\dfrac{12!}{\left( 12-2 \right)!2!}}=1-\dfrac{\dfrac{7!}{5!2!}}{\dfrac{12!}{10!2!}} \\
 & =1-\dfrac{7\times \dfrac{6}{2}}{12\times \dfrac{11}{2}}=1-\dfrac{7\times 3}{11\times 6} \\
 & =1-\dfrac{21}{66} \\
\end{align}\]
\[\therefore \] P (at least one black ball) \[=\dfrac{66-21}{66}=\dfrac{15}{22}\].