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An urn contains 2 red balls and 3 blue balls. Six drawing are made, the ball being placed back into the urn after each drawing. What is the probability, expressed as a rational number,that exactly 2 red balls are draw?

My solution

$\frac{10\times 5^4}{5^{6}}$ (10 from 5C2)

$\frac{2}{5}$

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  • $\begingroup$ took red fixed for first 2 slots then there are 10 such arrangements with 2 red balls $\endgroup$ – Tariro Manyika Mar 3 '19 at 4:02
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This is an example of a problem that can be solved via knowledge of the binomial distribution.

The probability of drawing a red ball is $\frac{2}{5}$. The probability of drawing a blue ball is $\frac{3}{5}$. There are $6$ drawings to be made (with replacement) and we are interested in the probability that exactly two of them will be red.

In other words, $n=6$, $k=2$ and $p=\frac{2}{5}$.

The probability of $k$ "successes" out of $n$ trials where probability of success if $p$ is:

$$Pr(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$$

Plugging in the appropriate values of $n,k,p$ give you the probability.

$\binom{6}{2}\left(\frac{2}{5}\right)^2\left(\frac{3}{5}\right)^4$


"Took red fixed for first two slots then there are 10 such arrangements with 2 red balls"

"My solution: $\frac{10\times 5^4}{5^6}$"

I do not understand why you thought this would work. If you fix the first two slots as red and then have no restriction on the remaining slots, what is to prevent you from having more than two reds? I further do not see how the expression you wrote has anything to do with your proposed solution.

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