A box containing 3 red balls and 2 white balls. A ball is taken randomly and with replacement. This is done 3 times.
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What is the probability that exactly one red ball is obtained?
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Probability at least 1 red ball is obtained?
Attempt
For the first part, the possibilities are: red ball at 1st draw, red ball at 2nd draw, or red ball at 3rd draw. The probabiity for each is $p = \frac{3}{5} \frac{2}{5} \frac{2}{5}$, so the answer is $3 \times p$.
For the second part, possibilities are: 1 red ball taken, 2 red balls taken, or 3 red balls taken. probability of 2 red balls taken is: $3 \times \frac{3}{5} \frac{3}{5} \frac{2}{5}$. So answer is:
$$ \left( 3 \times \frac{3}{5} \frac{2}{5} \frac{2}{5} \right) + \left( 3 \times \frac{3}{5} \frac{3}{5} \frac{2}{5} \right) + \frac{3}{5}\frac{3}{5}\frac{3}{5} $$
Best Answer
Both parts are correct. As regards the second one, it is easier to evaluate the complement: the probability that at least one red ball is obtained is $1$ minus the probability that all the three balls are white, namely $$1-\left(\frac{2}{5}\right)^3.$$