To find probability of an event Suppose three friends are asked to choose a ball blindly and with replacement from an urn containing each of the following colors red blue yellow green orange purple black. What is the probability  


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*all will draw same color?

*at least two will draw same color?                               


I tried this quetion as
First probability of choosing first friend $P(f_1)= \frac13$ similarly for other two friends $P(f_2)=\frac13$ and $P(f_3)= \frac13$ when first is chosen $P(\text{for first friend drawing a ball} )= \frac17$ similarly for other two friends. But how would I calculate that ball drawn is of same color?? And which color to choose or should I do for individual colour 
 A: You are not choosing a friend. There is a 100% chance that every friend will choose a ball, not a $\frac13$ chance. 
Let's first describe the probability space (the set of all possible outcomes and the probability of it occurring). Each friend will wind up with a color for a ball. There are seven possible colors that each friend will wind up with. Because the balls are replaced, what one friend chooses does not affect the choices available to a different friend. So, the total number of outcomes is the product of the choices for the individual friends (this is known as the Product Rule). 
Total number of choices: $7\times 7 \times 7 = 7^3$.
And every outcome is equally probable (there is one way out of the $7^3$ ways for friend 1 to get red, friend 2 to get red, and friend 3 to get red, just as there is one way out of $7^3$ for any other single outcome of friends choosing colors).
Now, we have two events. The first is that all three choose the same color. Since there are seven colors, choose one in $_{7}C_1 = \dbinom{7}{1} = 7$ ways. For each color chosen, you would need all three friends to choose that color. That is: $$\dfrac{7}{7^3} = \dfrac{1}{49}$$
For at least two friends choose the same color, the only way that will not happen is if every friend chooses a distinct color. The first friend has seven choices, the second has six, and the third has five. Thus, the probability for no friends choosing the same color is:
$$\dfrac{7\times 6\times 5}{7^3}$$
The probability that at least two friends choose the same color is the complement of that:
$$1-\dfrac{7\times 6 \times 5}{7^3} = \dfrac{19}{49}$$
