Question

An urn contains 9 red, 7 white and 4 black balls. If two balls are drawn at random. Find the probability that
(I) Both the balls are red.
(II) One ball is white
(III) The balls are the same colour.
(IV) One is white and the other red.

Answer 1

Hint: Here the given question is based on the concept of probability. We have to find the probability of following events. So, we have to write the given data in the numeral form and by definition of probability and using a formula of combination i.e., \[^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\], we write the solution for the given question.

Complete step by step solution:
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e., how likely they are to happen, using it. Probability can range in from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event.
The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.
\[\text{Probability of event to happen} P\left( E \right) = \dfrac{{\text{Number of favourable outcomes}}}{{\text{Total Number of outcomes}}}\]
Now consider the given question and we write the given data, so we have
Number of red balls = 9
Number of white balls = 7
Number of black balls = 4
Now, the total number of balls that urn contains = \[9\,red + 7\,white + 4\,black = 10\,balls\].
Two balls are drawn at random.
Therefore, the total number of outcomes \[n\left( S \right) = {\,^{20}}{C_2}\].

(I) Event A: Both the balls are red, then \[n\left( A \right) = {\,^9}{C_2}\]
The probability of random drawn balls both are red is
\[ \Rightarrow \,\,\,\,P\left( {red\,ball} \right) = \,\dfrac{{^9{C_2}}}{{^{20}{C_2}}}\]
The value of
\[^9{C_2} = \dfrac{{9!}}{{\left( {9 - 2} \right)!2!}} = \dfrac{{9 \times 8 \times 7!}}{{7!\, \times 2}} = 9 \times 4 = 36\] and
\[^{20}{C_2} = \dfrac{{20!}}{{\left( {20 - 2} \right)!2!}} = \dfrac{{20 \times 19 \times 18!}}{{18!\, \times 2}} = 10 \times 19 = 190\]. Then,
\[ \Rightarrow \,\,\,\,P\left( {red\,ball} \right) = \,\dfrac{{36}}{{190}}\,\,\,ways\]

(II) Event B: One ball is white.
The probability of random drawn balls which one ball should be white another will be red or black, then
\[ \Rightarrow \,\,P\left( {one\,white\,ball} \right) = \,\dfrac{{^7{C_1}{ \times ^{13}}{C_1}}}{{^{20}{C_2}}}\]
The value of
\[^7{C_1} = \dfrac{{7!}}{{\left( {7 - 1} \right)!1!}} = \dfrac{{7 \times 6!}}{{6!\, \times 1}} = 7\] and
\[^{13}{C_1} = \dfrac{{13!}}{{\left( {13 - 1} \right)!1!}} = \dfrac{{13 \times 12!}}{{12!\, \times 1}} = 13\]. Then,
\[ \Rightarrow \,\,P\left( {one\,white\,ball} \right) = \,\dfrac{{7 \times 13}}{{190}}\]
\[ \Rightarrow \,\,P\left( {one\,white\,ball} \right) = \,\dfrac{{91}}{{190}}\,\,ways\]

(III) The balls are the same colour.
The probability of random drawn balls both are either red, white and black colour the two balls will be the same colour.
\[ \Rightarrow \,\,\,P\left( {both\,balls\,of\,same\,colour} \right) = \,\dfrac{{^9{C_2}{ + ^7}{C_2}{ + ^4}{C_2}}}{{^{20}{C_2}}}\]
The value of
\[^7{C_2} = \dfrac{{7!}}{{\left( {7 - 2} \right)!2!}} = \dfrac{{7 \times 6 \times 5!}}{{5!\, \times 2}} = 7 \times 3 = 21\] and
\[^4{C_2} = \dfrac{{4!}}{{\left( {4 - 2} \right)!2!}} = \dfrac{{4 \times 3 \times 2!}}{{2!\, \times 2}} = 2 \times 3 = 6\], then
\[ \Rightarrow \,\,\,P\left( {both\,balls\,of\,same\,colour} \right) = \,\dfrac{{36 + 21 + 6}}{{190}}\]
\[ \Rightarrow \,\,\,P\left( {both\,balls\,of\,same\,colour} \right) = \,\dfrac{{63}}{{190}}\,\,ways\]

(IV) One is white and the other red.
The probability of random drawn balls on should be white and another should be red colour.
\[ \Rightarrow \,\,\,P\left( {one\,white\,and\,one\,red} \right) = \,\dfrac{{^9{C_1}{ \times ^7}{C_1}}}{{^{20}{C_2}}}\]
The value of
\[^9{C_1} = \dfrac{{9!}}{{\left( {9 - 1} \right)!1!}} = \dfrac{{9 \times 8!}}{{8!\, \times 1}} = 9\], then
\[ \Rightarrow \,\,\,P\left( {one\,white\,and\,one\,red} \right) = \,\dfrac{{9 \times 7}}{{190}}\]
\[ \Rightarrow \,\,\,P\left( {one\,white\,and\,one\,red} \right) = \,\dfrac{{63}}{{190}}\,\,ways\].
Hence, it’s a required solution.

Note:
The probability is a number of possible values. Must know we have to use the permutation concept or combination concept to solve the given problem because it is the first and main thing to solve the problem. Here we arrange the things in the possible ways so we are using the concept permutation.