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An urn contains $5$ red, $6$ blue and $8$ green balls. $3$ balls are randomly selected from the urn, find the probability of getting exactly one red ball if the balls are drawn with replacement.

Source: doubt came from the similar question here.

In the same line, my answer follows

Total number of ways $=19^3=6859$

Favorable ways $= 5×14×14+14×5×14+14×14×5=2940$

Probability $=\dfrac{2940}{6859}$

Is my understanding right? please correct if I am wrong.

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    $\begingroup$ I have not done the multiplication. But the structure of your answer is correct. The probability is $\frac{(3)(5)(14^2)}{19^3}$. $\endgroup$ – André Nicolas Jun 9 '16 at 21:11
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    $\begingroup$ this is correct $\endgroup$ – gt6989b Jun 9 '16 at 21:12
  • $\begingroup$ use Binomial distribution $\endgroup$ – Alex Jun 9 '16 at 23:10
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Your answer works. Another way to solve this is to note that because draws are done with replacement, the draws are independent and identically distributed. Thus the random variable $X=\text{number of red balls chosen in 3 draws}$ is a binomial random variable with $n=3$ and $p=\text{probability of red ball on a given draw}=\frac{5}{5+6+8}=\frac{5}{19}$. Thus

$$P(X=1)=\binom{3}{1}\cdot \left(\frac{5}{19}\right)^1\cdot \left(1-\frac{5}{19}\right)^2= \frac{2940}{6859}.$$

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Total number of ways $=19^3=6859$

Favorable ways $= 5×14×14+14×5×14+14×14×5=2940$

Probability $=\dfrac{2940}{6859}$

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  • $\begingroup$ Sine my approach and answer was correct, adding it as an answer here. $\endgroup$ – Kiran Jun 9 '16 at 21:15

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