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A bag contains \[8\]red and \[7\]black balls. Two balls are drawn at random. Find the probability that the both balls are of the same color.


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Hint: Probability is simply how likely something is to happen. Whenever we are unsure about the outcome of an event, we can talk about the probabilities of certain outcomes and how likely they are.
Probability of an event is a number between \[0\]& \[1\].
Probability of an event is given below by :
\[Probability\, = \,\dfrac{{Possible\,outcomes}}{{Total\,outcomes}}\].
In such questions for a solution we will use combinations.

Complete step by step solution:
Given,
A bag containing \[8\]red \[7\]black balls.
No. of balls drawn from the bag \[ = \,2\].
Total number of balls \[ = \]Red balls \[ + \]Black balls.
\[ = \,\,8\, + \,7\]
\[ = \,15\]balls.
Two balls are drawn randomly. We need to find the probability that both balls are of the same color i.e. either both are red or both are black.
We have,
Number of ways of drawing \[2\]balls from \[15\]balls.
\[ = \,{}^{15}{C_2}\]
\[\left[ {{}^n{C_r} = \dfrac{{n\,!}}{{r\,!\left( {n - r} \right)\,!}}} \right]\]
\[ = {}^{15}{C_2} = \dfrac{{15\,!}}{{2\,!\left( {15 - 2} \right)\,!}} = \dfrac{{15\,!}}{{2\,!\,\, \times 13\,!}}\]
\[ = \,\dfrac{{15 \times 14 \times 13!}}{{2 \times 1 \times 13\,!}}\] \[\left[ {\because n\,!\, = \,n\, \times \,\left( {n - 1} \right) \times x\left( {n - 2} \right)......} \right]\]
\[ = \,\dfrac{{15 \times 14}}{2} = 15 \times 7 = 105\]ways. _____(1).
First case:
NO. of ways of drawing \[2\]red balls \[ = {}^8{C_2}\]
\[ = \dfrac{{8!}}{{2!\left( {8 - 2} \right)!}}\]
\[\left[ {{}^n{C_r} = \dfrac{{n\,!}}{{r\,!\left( {n - r} \right)\,!}}} \right]\]
\[ = \dfrac{{8!}}{{2! \times 6!}} = \dfrac{{8 \times 7 \times 6!}}{{2 \times 1 \times 6!}}\]
\[ = \dfrac{{8 \times 7}}{2} = 4 \times 7 = 28\]ways. _________ (2).
Second case:
No. of ways of drawing \[2\]black balls \[ = {}^7{C_2}\]
\[ = \dfrac{{7!}}{{2!\left( {7 - 2!} \right)}}\]
\[\left[ {{}^n{C_r} = \dfrac{{n\,!}}{{r\,!\left( {n - r} \right)\,!}}} \right]\]
\[ = \dfrac{{7!}}{{2! \times 5!}} = \dfrac{{7 \times 6 \times 5!}}{{2 \times 1 \times 5!}}\]
\[ = \dfrac{{7 \times 6}}{2} = 7 \times 3 = 21\]ways. _________ (3).
Hence, Probability of drawing balls of same color
\[ = \dfrac{{Probability\,that\,both\,are\,red + Probability\,that\,both\,are\,black}}
{{Probability\,of\,drawing\,2\,balls\,at\,all\,from\,total}}\]
\[ = \dfrac{{28 + 21}}{{105}}\] [from equations 1, 2, & 3]
\[ = \dfrac{{49}}{{105}}\]\[ = \dfrac{7}{{15}}\].

Note: We see combinations when we need to select things.
When selecting/choosing something never confuse it with permutation
Solution of r things from n things is \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\].