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A bag contains 15 balls, of which 5 balls are red, 5 balls are green, and 5 balls are blue.

We are given 3 attempts to draw a ball from this bag randomly, after each attempt, the drawn ball would be kept back in the bag so during each attempt the total balls in the bag would always be 15. My question is, what is the probability of drawing atleast 1 red ball in 3 attempts.

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    $\begingroup$ This is a classical binomial distribution. $\endgroup$ – Masacroso Nov 22 '15 at 9:57
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The probability of drawing a non-red ball from one attempt is $\frac{10}{15} = \frac{2}{3}$.
Thus the probability of drawing only non-red balls from three attempts is $(\frac{2}{3})^3 = \frac{8}{27}$.

Therefore the probability of drawing at least one red ball from three attempts is:

$$1 - \frac{8}{27} = \frac{19}{27}$$

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  • $\begingroup$ Only one question, what would be the new result if the condition that the balls in the bag remain same is removed? $\endgroup$ – Sarthak123 Nov 22 '15 at 10:13
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    $\begingroup$ $1 - \frac{10}{15}\frac{9}{14}\frac{8}{13}$ $\endgroup$ – Petar Ivanov Nov 22 '15 at 10:14

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