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"A box contains 2 red balls and 4 yellow balls. If 2 balls are randomly chosen and simultaneously removed from the box, what is the probability that only yellow balls are left in the box?"

My work: The total possible outcomes of 2 balls removed are { 1 red 1 yellow, 2 red 0 yellow, 0 red 2 yellow }

Is it correct to say the required probability = 1/3?

Can any buddy help?

user256670
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  • Your reasoning is slightly askew. The outcome of, for example, "$1$ red, $1$ yellow" actually corresponds to $2 \times 4 = 8$ outcomes, because there are $2$ ways to choose a red ball, and $4$ ways to choose a yellow ball. – Colm Bhandal Jul 24 '15 at 17:54

4 Answers4

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Not correct. The outcomes you list are not equi-probable.

Either multiply probabilities, Pr = $\frac26\cdot\frac15 = \frac{1}{15}$

or use combinations, Pr = ${2\choose 2}\div{6\choose 2} = \frac{1}{15}$

true blue anil
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  • sorry for stupid question again, why using multiply and not using add? Please enlighten me – user256670 Jul 24 '15 at 18:01
  • Very briefly, multiplication for AND, addition for OR. You can see in more detail at https://people.richland.edu/james/lecture/m170/ch05-rul.html – true blue anil Jul 24 '15 at 18:41
  • Yes, in textbook it says similar points now you have summarized a bit for me. And thanks for the link. – user256670 Jul 24 '15 at 18:44
  • If I understand correctly, then quoting from the link: "Dependent Events

    If the occurrence of one event does affect the probability of the other occurring, then the events are dependent.

    Conditional Probability

    The probability of event B occurring that event A has already occurred is read "the probability of B given A" and is written: P(B|A)

    General Multiplication Rule

    Always works.

    P(A and B) = P(A) * P(B|A)" Am I correct?

     – user256670 Jul 24 '15 at 18:47
  • Yes. P(1st red) = 2/6, P(2nd red | 1st red) = 1/5 – true blue anil Jul 24 '15 at 18:50
  • You're welcome. Study well the basic principles ! – true blue anil Jul 24 '15 at 19:48
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The total number of outcomes is the total number of pairs of balls. There are $6$ balls, so there are $6 \choose 2$ pairs. Only one of these pairs is $(red, red)$.

Colm Bhandal
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  • Colm, thanks, for the help, since I am still not yet familiar with the combination and permutation chapters yet so I get a bit lost, but I will think it and try to understand it, – user256670 Jul 24 '15 at 18:15
  • No problem. Combinations, permutations and factorials are the bread and butter of combinatorics, so well worth learning. And even deeper is a really good understanding of the meaning of multiplication. – Colm Bhandal Jul 24 '15 at 18:17
  • I am just confused by 'when to use multiplication', 'when to use addition'. The most difficult aspect of doing probability exercise is that I don't know how to verify my answer, compared to other topics like geometry, percentage and algebra. – user256670 Jul 24 '15 at 18:32
  • My advice is: spend some time delving into the meanings of multiplication and addition. I'd advise you to read @anakhronizein's answer in the following link: https://math.stackexchange.com/questions/1369974/how-many-words-can-be-formed-using-all-the-letters-of-daughter-so-that-vowels/1369982#1369982. – Colm Bhandal Jul 25 '15 at 11:25
  • As for checking your answer, it is hard in general, but there are workarounds. In this case, because the space of total possible outcomes is small ($6 \choose 2 = 15$) you could actually draw all 15 possibilities out. Label your red balls $r_1, r_2$ and your yellow balls $y_1, y_2, y_3, y_4$ and then write every pair e.g. ${r_2, y_3}$ exactly once- ignoring ordering. Then you will see that there are $15$ total pairs and ${r1, r2}$ appears only once. – Colm Bhandal Jul 25 '15 at 11:31
  • You could also check your answer by experiment. Get hold of $6$ balls- $2$ red and $4$ yellow. Or use different objects/colours. Put them in a bag. Close your eyes. Draw two out at random. Record your result. Do this enough times and your should (if it's truly random) see the combination of two reds about $1/15^{th}$ of the time.

    Finally, particularly for bigger state spaces, you could write a computer program to simulate an experiment. This would require knowledge of programming though so I'm not sure if that's a viable option here.

     – Colm Bhandal Jul 25 '15 at 11:33
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The probability of picking a yellow in $2$ goes is:

$P(\text{pick a yellow}) + P(\text{pick a red then a yellow})\\ = \dfrac{4}{6} + \dfrac{2}{6}\cdot\dfrac{4}{5}\\ =\dfrac{2}{3} + \dfrac{1}{3}\cdot\dfrac{4}{5}\\ =\dfrac{10}{15} + \dfrac{4}{15}\\ =\dfrac{14}{15}$

So the probability of not picking a yellow, i.e. pick both reds, is $1-P(\text{pick a yellow in 2 goes})=1-\dfrac{14}{15}=\dfrac{1}{15}$

JMP
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  • Jon, great explanation. That's the next thing I would like to ask: because in the question it says removed simultaneously, so is that different from 2 goes? I am just a bit confused. – user256670 Jul 24 '15 at 18:16
  • @user256670; it doesn't really mean anything, its a red herring – JMP Jul 24 '15 at 18:18
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If only yellow balls remain, the $2$ that you picked are the $2$ red ones. We are simultaneously removing the two balls, so we can view this as 'not replacing'.

What is the probability of picking a red one on the first of the $2$ that you draw? That is $\dfrac26 = \dfrac13$. What about picking a red on the second of the $2$? Well now we have $5$ balls to choose from and only $1$ is red. So the probability is $\dfrac15$. $$\frac13 \cdot \frac15 = \boxed{\frac{1}{15}}$$

Rick
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  • but one thing i am still not very clear about is that if we calculate the 1st ball's probability, then followed by the 2nd one, then we are not complying to the question because we are separating the two event? The question says simultaneously meaning taking out 2 in 1 go...? ...confused ... – user256670 Jul 24 '15 at 18:57
  • @user256670. If we are simultaneously doing it, it actually equal to the probability of doing them separately and with out replacement. – Rick Jul 24 '15 at 19:03
  • Ok, I think I get it buddy. I guess our answer should be correct, because, the no of red ball is smaller than the number of yellow balls, so the chances of red ball totally got removed should be a very small number. 1/15 should be just about right. – user256670 Jul 24 '15 at 19:09