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Question: A bag contains 4 white, 7 black and 5 red balls. 4 balls are drawn with replacement. Show that the probability of at least two balls are white is $\frac{67}{256}$.*

I have tried in the following way:

Number of white balls = 4
Number of black balls = 7
Number of red balls = 5
Total balls = 16

$P (\text{at least $2$ balls drawn are white})$ $= P (2\text{ balls drawn are white}) + P (3 \text{ balls drawn are white}) +P (4 \text{ balls drawn are white}) $ $=\frac{^4C_2\times ^{12}C_2}{^{16}C_4}+\frac{^4C_3\times ^{12}C_1}{^{16}C_4}+\frac{^4C_4\times ^{12}C_0}{^{16}C_4}$

I don't know whether my approach is correct or not.

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  • $\begingroup$ It looks like you may not have accounted for the fact that, using your first term as an example, you also require the other two balls to be non-white. What does that calculation give you? $\endgroup$ – The Count Jan 18 '17 at 3:08
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Note you're sampling without replacement, so each ball drawn can be treated as an independent trial with $(4/16)$ chance of drawing a white ball.

If two are white, any two of the four can be white, so there are ${4\choose 2 } = 6$ possibilities, each of which have probability $(4/16)^2(12/16)^2$ so the probability that two are white is $6*(4/16)^2*(12/16)^2.$

Similarly if three are white, there are $4$ ways and the probability is $$4*(4/16)^3*(12/16)^1.$$

And the probability all four are white is $(4/16)^4.$

The total probability that two or more are white is the sum $$\frac{6*4^2*12^2 + 4*4^3*12 + 4^4}{16^4} = \frac{17152}{16^4} = \frac{67}{256} $$

Your approach of using $12 \choose 2$, etc is better suited to sampling without replacement.

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Let X be number of white balls during 4 draws, we can see X follows Binomial distribution because of drawing with replacement: Bin(4, 1/4).

So $P(X=k) = {4 \choose k}\frac{1}{4}^k\frac{3}{4}^{4-k} $

$ P(X\ge2) = 1 - P(X=0) - P(X=1) $

Plug in 0 and 1 to the PMF and get the answer $\frac{67}{256}$.

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Well, the probability that any one of them will be white is $1/8$, correct? So if you want at least two of them to be white, then you can find the probability that only one or zero of them will be white, then take the inverse of that.

$$\begin{align*}P(\text{one ball will be white}) &= \frac18\cdot\left(\frac78\right)^3\cdot4=\frac12\cdot\left(\frac78\right)^3 \\ P(\text{no balls will be white}) &= \left(\frac78\right)^4\\ P(\text{at least two will be white})&=1-P(\text{one will be white}) - P(\text{no balls will be white})\end{align*}$$

Calculate the rest from there.

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To follow a more systematic approach that you can use in general, start with $$ p_{\;1} = {4 \over {16}}\quad p_{\;2} = {7 \over {16}}\quad p_{\;3} = {5 \over {16}} $$

Because draws are with replacement, they are constant at each draw.

Now, consider that $$ 1 = p_{\;1} + p_{\;2} + p_{\;3} = \left( {p_{\;1} + p_{\;2} + p_{\;3} } \right)^{\,4} = \left( {p_{\;1} + \left( {1 - p_{\;1} } \right)} \right)^{\,4} $$

From this, applying the binomial (or multinomial) expansion you can compute all the concerned probabilities.

In your case $$ P\left( {2 \le w} \right) = \sum\limits_{2\, \le \,k\, \le 4} {\left( \matrix{ 4 \cr k \cr} \right)p_{\;1} ^{\,k} \left( {1 - p_{\;1} } \right)^{\,4 - k} } = {{67} \over {256}} $$

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