There are $6$ red balls and $5$ blue balls in a jar. You pick any $4$ balls without looking in the jar. What is the probability that you would be having $4$ red balls in hand?

Note that you're picking up all the $4$ balls in one single attempt and not one-by-one. Also balls of same colour are to be considered as identical. I tried this and got the answer as $4/11$ which was wrong. I can't figure out what the cases (sample space) would be.

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    $\begingroup$ If you grab $4$ balls then one of them will somehow be elected as "first ball touched". After that a ball is labeled as "second ball touched", etc. This is an effort to clarify that there is no difference between picking up in a single attempt or one by one. $\endgroup$
    – drhab
    Mar 8, 2016 at 13:57
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    $\begingroup$ Reminds me of a Dr. Deming lecture I once attended. He has volunteers from the audience pick colored balls from a bowl held up high (so they can't see...) He rewards some folks for high returns, and berates, yells and really belittles folks who perform poorly. Sitting in the audience I really cringed. The results are random. It really makes the point that we as managers are total idiots. $\endgroup$
    – zipzit
    Mar 8, 2016 at 16:56
  • $\begingroup$ As an aside, this problem could be reworded as: "There are 11 balls in a jar, numbered 1-11. You pick out four balls at random. What is the probability that all four balls will be odd numbers?" $\endgroup$
    – J.R.
    Mar 8, 2016 at 20:23
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    $\begingroup$ In the future, consider reading the textbook rather than asking us to do your homework for you. $\endgroup$
    – jfa
    Mar 9, 2016 at 14:46

4 Answers 4


You are picking 4 balls without replacement. The probability that you pick four red balls is $$P =\frac{6}{11} \cdot \frac{5}{10}\cdot \frac{4}{9}\cdot \frac{3}{8}\\P=\frac{360}{7920}\\P=\frac{1}{22}\\P \approx 0.045 \\P \approx 4.5\%$$

  • $\begingroup$ We are picking all the 4 balls at the same time . $\endgroup$
    – Tejus
    Mar 8, 2016 at 13:47
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    $\begingroup$ @Tejus that makes no essential difference $\endgroup$
    – drhab
    Mar 8, 2016 at 13:48
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    $\begingroup$ @Tejus Pick the balls at the same time. Now we will check whether they have the color red. You take a look at each ball. What is probability that the first ball you look at is red? $\frac6{11}$. If that appears to be the case, then what is the probability that the second ball looked at also will appear to be red? $\frac5{10}$. Et cetera. $\endgroup$
    – drhab
    Mar 8, 2016 at 14:59
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    $\begingroup$ @Tejus - The only way the numbers would be different by selecting the balls one-at-a-time would be if the selected balls were returned to the bin after each pick. If they are kept out of the hopper, though, then it makes no difference if you pull them out one at a time, or all at once. $\endgroup$
    – J.R.
    Mar 8, 2016 at 18:54
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    $\begingroup$ @Tejus -- The equation in J.Bush's answer isn't showing you sequential picks per say. It's showing how to calculate the number of ways 4 Red (and only Red) balls can be picked over the total number of ways in which 4 balls (of any color and sequence) can be chosen. -- Both the numerator and denominator decrease by one proportionally to the number of balls selected. One way to show this is to specify the number of subjects involved as if one ball were being selected iteratively four times. --- He could have started with P = 360/7920 but that only assumes the first step anyway. $\endgroup$
    – user23715
    Mar 8, 2016 at 21:51

Since you have to pick 4 balls, the number of elements in the sample space would be $^{11}C_4$ ("11 choose 4"). The number of ways of picking all red balls would be $^6C_4$.

Therefore, the probability would be $$={{^6C_4}\over{^{11}C_4}} \\ ={1\over 22}$$

  • $\begingroup$ Balls if the same colour are identical $\endgroup$
    – Tejus
    Mar 8, 2016 at 13:49
  • $\begingroup$ @Tejus, thats why the 6 choose 4 are all considered the same outcome. $\endgroup$
    – Octopus
    Mar 8, 2016 at 18:57

The answer is easly given considering permutations. Indeed the probability is: $$P =\frac{7!}{11!} \cdot \frac{6!}{2!}=\frac{1}{22}$$ where

  • $11$ is the total number of balls;
  • $7$ the number of balls remaining in the jar;
  • $6$ the number of red balls;
  • $2$ the number of red balls remaining in the jar.
  • $\begingroup$ Ah yes, the Factorial method! Excellent way to show the balls are all chosen "at once" and not sequentially. For the OP this might be the best (read - most helpful) answer. $\endgroup$
    – user23715
    Mar 8, 2016 at 21:54

Well its very simple.

See first lets calculate the sample space. It would be $$ {^{11}C_4} $$ because you are choosing a total of 4 balls out of 11 balls.

Next you need to calculate the favorable event

so you have 6 blue balls and favorable would be when you take out 4 blue ones. which can be done by $$ {^6C_4} $$ ways. So the probability becomes $$ ={{^6C_4}\over{^{11}C_4}} \\={1\over 22} $$ If you dont get it by this method here is the basic step by step trick:- Probability of getting first blue ball is $$ ={6\over 11} $$ as there are total 11 and 6 blue now Probability of getting Second blue ball is $$ ={5\over 10} $$ because you took out 1 ball from total which is blue so both numbers reduce by 1. similarly 3rd blue ball $$ ={4\over 9} $$

and 4th one $$ ={3\over 8} $$ so final probablity is $$ P =\frac{6}{11} \cdot \frac{5}{10}\cdot \frac{4}{9}\cdot \frac{3}{8}\\P=\frac{360}{7920}\\P=\frac{1}{22}\\ $$

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    $\begingroup$ Please avoid saying that a problem is simple. If the person who posted the question found it simple, she or he would not have asked the question in the first place. $\endgroup$ Mar 8, 2016 at 15:51
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    $\begingroup$ @Taussig, I took Kunal's statement as meaning that the answer can be found by following a simple procedure (if you know the procedure), not that the question is simple. But I concede it could easily be taken the way you read it. $\endgroup$
    – LarsH
    Mar 8, 2016 at 17:10

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