A box contains 12 balls numbered 1 through 12. If 5 balls are selected one at a time from the box, without replacement, what is the probability that the largest number selected will be 9?

i want just say $\frac{9*8*7*6*5}{12P5}$ but that is wrong and dont know why.

Anonymous 06/12/2018 at 22:04.

The event "the largest draw is $9$" can be seen as the intersection of two events:

- $E_1$: all draws are at most $9$.
- $E_2$: (at least) one draw is $9$.

So you can compute your probability as $P(E_1)\cdot P(E_2|E_1)$.

$P(E_1)={9\choose 5} / {12 \choose 5} = \frac{7}{44}$, i.e. the number of ways you can take your $5$ draws from the $[1...9]$ interval, over the number of ways you can take them from the $[1...12]$ interval.

$P(E_2|E_1)=1-\frac{8}{9}\frac{7}{8}\frac{6}{7}\frac{5}{6}\frac{4}{5}=\frac{5}{9}$, i.e. $1$ minus the probability of avoiding a $9$ with your first, second, third, fourth and fifth draw *given that all your draws are in the $[1...9]$ interval*.

Then the probability that your highest draw is a $9$ is $\frac{7}{44}\frac{5}{9}=\frac{35}{396}$. There are other, slightly quicker ways to compute the solution, but I think that this one through conditional probabilities is the most "obvious".

The original question also asks: why is the probability *not* $\frac{9\cdot 8\cdot 7\cdot 6\cdot 5}{12 \choose 5}$? The denominator is the number of ways one can choose $5$ balls out of $12$. The numerator is the number of ways one can choose $5$ "distinguishable" balls out of $9$. This has two problems: first, it counts the balls as "distinguishable" (in other words, it takes $9\cdot 8\cdot 7\cdot 6\cdot 5$ instead of $9\choose 5$). Second, it considers every case in which the highest draw is *at most* $9$, rather than *exactly* $9$ (so it forgets to multiply by the probability that of the $5$ draws in the $[1-9]$ range, one is indeed $9$, i.e. $1-\frac{8}{9}\frac{7}{8}\frac{6}{7}\frac{5}{6}\frac{4}{5}=\frac{5}{9}$). Addressing these two issues takes us to the formula above.

herb steinberg 06/12/2018 at 21:29.

You need to do the problem in two parts. First what is the probability that none of the ball chosen are numbered 10 through 12. This has probability $P=\frac{9\times 8\times 7\times 6 \times 5}{12 \times 11\times 10\times 9\times 8}$. Second the probability (under this condition) that a 9 has been chosen. To get this calculate the probability that a 9 was not chosen. This is $Q=\frac{8\times 7\times 6\times 5\times 4}{9\times 8\times 7\times 6\times 5}$. The final answer is $P(1-Q)$

E-A 06/12/2018 at 21:28.

The event in which the largest number is 9 is if:

a) You pick 9

b) You pick 4 other numbers that are less than 9

The number of ways you can do that is simply picking 4 items out of 8. You should be able to finish from here.

Graham Kemp 06/13/2018 at 00:40.

A box contains 12 balls numbered 1 through 12. If 5 balls are selected one at a time from the box, without replacement, what is the probability that the largest number selected will be 9?

The probability for selecting `9`

*and* four numbers less than `9`

, when selected any five of the twelve without replacement, is: ${^8\mathrm C_4}\,/\,{^{12}\mathrm C_5}$ or $5\cdot{^8\mathrm P_4}\,/\,{^{12}\mathrm P_5}$

Since you seem more familiar with using $^n\mathrm P_r$ then that second is derved as follows: Count the ways to permute the maximum (`9`

) and four places. Count the ways to fill those places with a permutation of 4 balls from the 8 lesser balls. Count the ways to fill all five places with a permutation of any 5 balls from 12. Multiply and divide the counts, as appropriate.

i want just say $\frac{9*8*7*6*5}{12P5}$ but that is wrong and dont know why.

${^9\mathrm C_5}/{^{12}\mathrm C_5}$, or ${^9\mathrm P_5}/{^{12}\mathrm P_5}$, is the probability for selecting five numbers that are `9`

or less, when selected any five of the twelve without replacement. (The maximum of those five ballsis not restricted to being `9`

; just to not exceeding it.)

KonKan 06/13/2018 at 00:56.

There are $8$ choices (i.e.: 1,2,3,4,5,6,7,8) which are less than $9$, $1$ choice which equals $9$ and $3$ choices (i.e.: 10,11,12) which exceed $9$. So, appplying the hypergeometric distribution formula, we readily get: $$ \frac{\binom{8}{4}\binom{1}{1}\binom{3}{0}}{\binom{12}{5}}=\frac{\binom{8}{4}}{\binom{12}{5}}=\frac{\frac{5\cdot 6\cdot 7\cdot 8}{2\cdot 3\cdot 4}}{\frac{8\cdot 9\cdot 10\cdot 11\cdot 12}{2\cdot 3\cdot 4\cdot 5}}=\frac{5\cdot 6\cdot 7\cdot 8}{8\cdot 9\cdot 2\cdot 11\cdot 12}=\frac{5\cdot 7}{3\cdot 11\cdot 12}=\frac{35}{396} $$

cmitch 06/12/2018 at 21:42.

EDIT: Misread initial question at first. Probability that largest number is nine would be 5*(8 choose 4)/(12 choose 5). 8 choose 4 represents all the ways of choosing the remaining 4 numbers below 4 without replacement, but we multiply by 5 as well as for each combination of 4 there are 5 ways to include the number 9. Then divide by the total number of combinations

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