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A jar contains 5 marbles, 1 of each of the colors- Red, White, Blue, Green and Yellow.

If 4 marbles are removed from jar, what is the probability that yellow one was removed?

My attempt:

Each marble is equally likely to be removed. Hence the probability of selecting a marble of any color is 1/5

number of ways to select 4 marbles: 5C4

probability of selecting a yellow marble= 1/5

probability of selecting 3 other marbles = 1/5*1/5*1/5

Hence probability that the yellow marble is removed is 5C4 * 1/5* (1/5*1/5*1/5)

What is wrong with this approach?

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The probability yellow wasn't removed is clearly $\dfrac{1}{5}$. –  André Nicolas Dec 17 '12 at 19:09
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3 Answers

up vote 1 down vote accepted

The number of ways to select four marbles, one of which is yellow, would in this case be $${}_1C_1\cdot{}_4C_3=1\cdot 4=4,$$ so the probability of selecting the yellow marble is $$\frac{4}{{}_5C_4}=\frac45.$$

Alternately, we can proceed stepwise as follows: There's a $\frac45$ chance that the first marble isn't yellow. If the first marble isn't yellow, then there's a $\frac34$ chance that the second marble isn't yellow. If the first two marbles aren't yellow, then there's a $\frac23$ chance that the third marble isn't yellow. If the first three marbles aren't yellow, then there's a $\frac12$ chance that the fourth marble isn't yellow. Therefore, there's a $$\frac45\cdot\frac34\cdot\frac23\cdot\frac12=\frac15$$ chance that none of the four marbles drawn is yellow, so there's a $$1-\frac15=\frac45$$ chance that one of the four marbles is yellow.


As a third approach (which I'll discuss in more detail), since there's only one yellow marble, then to get the probability that the yellow marble was chosen, we need only add the probability of the following distinct events: (i) the yellow marble was chosen first, (ii) the yellow marble was chosen second, (iii) the yellow marble was chosen third, (iv) the yellow marble was chosen fourth. Hopefully, you see why these events have no overlap (mutually exclusive), and why together they comprise all the possible ways that the yellow marble could be chosen in this circumstance.

We already know that $$P(\text{yellow first})=\frac15.\tag{i}$$ If yellow is chosen second, then some other marble was chosen first--there are ${}_4C_1=4$ ways this can happen out of ${}_5C_1=5$ possibilities--leaving $1$ yellow marble out of a total of $4$ remaining, so $$P(\text{yellow second})=\frac45\cdot\frac14=\frac15.\tag{ii}$$ If yellow is chosen third, then two non-yellow marbles were chosen first--there are ${}_4C_2=6$ ways this can happen out of ${}_5C_2=10$ possibilities--leaving $1$ yellow marble out of a total of $3$ remaining, so $$P(\text{yellow third})=\frac{6}{10}\cdot\frac13=\frac15.\tag{iii}$$ If the yellow marble is chosen fourth, then three non-yellow marbles were chosen first--there are ${}_4C_3=4$ ways this can happen out of ${}_5C_3=10$ possibilities--leaving $1$ yellow marble out of a total of $2$ remaining, so $$P(\text{yellow fourth})=\frac{4}{10}\cdot\frac12=\frac15.\tag{iv}$$ Thus, $$\begin{align}P(\text{yellow chosen}) &= P(\text{yellow first})+P(\text{yellow second})+P(\text{yellow third})+P(\text{yellow fourth})\\ &= \frac15+\frac15+\frac15+\frac15\\ &= \frac45.\end{align}$$

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awesome explanation. thanks a lot. but can you please tell me what is wrong with my attempt? (selecting 4 marbles including a yellow and multiplying with their probabilities) –  Karan Dec 17 '12 at 19:18
    
If you're going to count the total number of possible outcomes, your next step is to count the number ways to achieve the desired outcome, then calculate the probability of the desired outcome by taking $\frac{\text{# desired}}{\text{# total}}$. (cont'd) –  Cameron Buie Dec 17 '12 at 19:33
    
You say that the probability of choosing a marble of any particular color is $\frac15$, but that isn't so. What you can say is that the probability of choosing a marble of any particular color as the first marble is $\frac15$. The probability then changes as we choose more marbles. I'll add a third alternative approach to my answer momentarily. –  Cameron Buie Dec 17 '12 at 19:36
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Probability of not selecting the yellow marble is $$\frac{\binom44}{\binom54}=\frac15$$

So, the required probability $$1-\frac15=\frac45$$

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the answer is 4/5 –  Karan Dec 17 '12 at 19:14
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Selecting 4 marbles to take out is the same as selecting one marble to not take out. There are 5 marbles you can chose. However you want to keep the yellow one in it. So there is only one out of the five combinations which satisfies this. Therefore the probability is 4/5 as André Nicolas pointed out.

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