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15 coupons are numbered 1,2,3...15. 7 coupons are selected at random one at a time with replacement. Find probability that 9 is the largest number appearing on the coupon. I am having a problem finding $N(a)$ for this problem. My working:

$N(s)= 15^7$ (since you have 15 choices for each time you pick up a coupon)

$N(a)= 9^7$ (since you have 9 permissible choices for each time you pickup a coupon)

But according to my textbook $N(a)= 9^7 -8^7$ but no further explanation is given. So i would really appreciate it if someone could point out where i am going wrong. Thanks in advance :)

AugSB
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Dèsjardins
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4 Answers4

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$9^7$ are the possibles 7-uples, but you want that at least one is 9. The competent of this is the 7-uples such that all elements aren't 9, there are $8^7$

Martín Vacas Vignolo
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The point is that $9^7$ is the ways in which you can obtain $7$ numbers between $1$ and $9$, but since $9$ must be the max, you should also avoid all the cases in which all the numbers are from $1$ to $8$, making the max smaller than $9$, hence $9^7-8^7$.

Giovanni Resta
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There are $8^7$ combinations in which largest number appearing on a coupon is at most $8$.

There are $9^7$ combinations in which largest number appearing on a coupon is at most $9$.

Hence there are $9^7-8^7$ combinations in which largest number appearing on a coupon is $9$.

barak manos
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Question: Fifteen Coupons are numbered 1 to 15. 7 coupons are selected at random, one at a time with Replacement. What is the Probability that the Largest number appearing on the selected coupon is 9 ???

Solution :

With Replacement => Binomial Distribution .

Success = p = Probability of getting a coupon Bearing number 9 is $\frac{1}{15}$

Failure =$(1-p)= q =$ Probability of getting a coupon DO NOT Bearing number 9 is $(1 - \frac{1}{15}) = \frac{14}{15}$

Out of Fifteen, $7$ coupons are randomly selected with Replacement. Therefore, $n = 7$ . Here r $= 1$ .

$P(X = r) = \binom{n}{r}p^{r}(1-p)^{n-r} = \binom{7}{1}.(\frac{1}{15})^{1}. (\frac{14}{15})^{6} = \frac{7}{15}.(\frac{14}{15})^{6}$

Soudipta Dutta
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