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This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with probability, which perhaps yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

A bag contains seven balls numbered from $1$ to $7$. A ball is chosen at random and its number is noted. The ball is then returned to the bag. This is done a total of seven times.

$(a)$ What is the probability that each ball is selected exactly once?

$(b)$ What is the probability that at least one ball is not selected?

$(c)$ What is the probability that exactly one of the balls is not selected?

My thoughts:

$(a)\, \left(\frac 17\right)^7=\frac{1}{823543}$

$(b) \, 1 - \frac{1}{823543}?$

$(c)$ No idea on this one. My head aches after just thinking about it.

Em.
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anonymous
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    The first one should be $\dfrac{7!}{7^7}$ (and the second one would be its complement then). Think about it. Or look at the smaller case of the same problem with $2$ instead of $7$ everywhere. – M. Vinay May 28 '16 at 03:12
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    Why do you say that it has "something to do with probability": it is visible that this is the case: the word "probability" appears in every question ! – Jean Marie May 28 '16 at 03:18

5 Answers5

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  1. There are $7!$ ways of picking $7$ different balls (in order), and $7^7$ ways of picking $7$ ones with possible repetition (in order). So the answer to the first question is $\dfrac{7!}{7^7}$.
  2. This is the complement of the first one, so the answer is $1 - \dfrac{7!}{7^7}$.
  3. There are $7$ ways of choosing the one ball that will not be included in the selection. Since we must include all the other six balls, and we have to pick a total of seven, our final selection will consist of seven balls with exactly one repeated. Again, there are $6$ ways of choosing the ball that will be repeated. Also, there are $\dfrac{7!}{2}$ ways of arranging these seven balls (in order). Thus, the numerator is $7 \times 6 \times 7!/2$, and the denominator, as before, is $7^7$. Therefore, the final answer is $\dfrac{7 \times 6 \times 7!}{2 \times 7^7} = \dfrac{3 \times 6!}{7^5}$.
M. Vinay
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What is the probability that each ball is selected exactly once?

$$\frac{7!}{7^7}$$

What is the probability that at least one ball is not selected?

$$1-\frac{7!}{7^7}$$

What is the probability that exactly one ball is not selected?

$$\frac{\binom{7}{6}\cdot\binom{6}{1}\cdot\frac{7!}{1!\cdot1!\cdot1!\cdot1!\cdot1!\cdot2!}}{7^7}$$

barak manos
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  • @AndréNicolas: Yes, I agree, I did not take into account which ball will be selected twice. Thank you for pointing this out, I will modify or delete :) – barak manos May 28 '16 at 03:25
  • @AndréNicolas: Yes, my clumsy-ness!!! I meant to say $\frac{7!}{1!1!1!1!1!2!}$ – barak manos May 28 '16 at 04:02
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(a) - you need the probability at each step of the next ball being none of hte preceeding balls. Only when you get to the last ball is that as small as $\frac 17$.

$$ 1.\frac 67.\frac 57.\frac 47.\frac 37.\frac 27.\frac 17 = \frac{7!}{7^7} \approx 0.00612$$

(b) As you correctly infer, this is $1-$ the result above. $$1-0.00612 = 0.99888$$

(c) This is better handled with a little more structure. Obviously there are $7$ choices for which ball is missed, so the probability is $7$ times the probability that only (say) ball $\#1$ is missed. Then if ball $\#1$ is missed, one of the other balls is duplicate - now the total probability is multiplied by 6 from the specific case "ball $\#1$ is missed, ball $\#2$ is duplicated". Now we need the distinct arrangements of $\{2234567\}$, which is $\frac {7!}{2!}$. Total arrangements for the one ball missing case is then: $$ 7 \times 6 \times \frac {7!}{2!} = 105840$$ which can then be compared to the selection universe of $7^7 = 823543$ t give a probability of $$ \frac {105840}{823543} \approx 0.128518 $$

Joffan
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a. Each time we pick one of the seven balls, so there are totally $7^7$ ways to do it. If each ball is selected exactly ones, it is equivalent to permutating the numbers $\{1,2,3,4,5,6,7\}$. For example, if the permutation is $\{1,3,5,4,2,7,6\}$, then the ball labelled by $'1'$ is selected at the first time, the ball labelled by $'3'$ is selected at the second time, $\dots$, the ball labelled by $'6'$ is selected at the last time. So, there are $7!$ ways such that each ball is selected exactly once. The probability is

$$\frac{7!}{7^7}=\frac{720}{117649}.$$

b. You have the right idea.

$$1-\frac{7!}{7^7}=\frac{116929}{117649}.$$

c. Assume the ball $'1'$ is not selected, then each of the six balls $'2','3','4','5','6','7'$ is selected during the seven times. Thus, exactly one of them are selected twice and the rest are selected exactly once. If the ball $'2'$ is selected twice, then it is equivalent to a permutation of $\{2,2,3,4,5,6,7\}$. There are $\frac{7!}{2}$ ways to do it. (We divide $7!$ by $2$ because there are two $'2'$ here, for each permeation if we switch the two $'2'$, the permutation is still the same.) The situations are the same for the ball $'i'$ is not selected and the ball $'j'$ is selected twice.($1\leqslant i,j\leqslant 7$ and $i\neq j$). Thus, there are totally $7\times 6\times \frac{7!}{2}$ ways to do it

$$\frac{7\times 6\times \frac{7!}{2}}{7^7}=\frac{2160}{16807}.$$

Qingzhong Liang
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Ad (a): You have calculated the probability that a specific order is selected. But there are $7!$ possible orders.

Ad (b): You are right to use the converse probability. Just use the hint for (a).

Ad (c): The result of the selected balls look like $AABCDEF$. It can be arranged in $\frac{7!}{2!}=21$ ways. $AA$ can be $one$ out of $seven$. Given $AA$, $BCDEF$ has to be $five $ out of $six$. In total the probability is $\frac{7!\cdot 7\cdot 6}{2!\cdot 7^7}=\frac{3\cdot 6!}{7^5}\approx 12.85\%$

callculus42
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