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Fifteen coupons are numbered 1 to 15. Seven coupons are selected at random, one at a time with replacement. What is the probability that the largest number appearing on a selected coupon be 9?

My attempt: I tried to break down the sum into 7 parts and find the probability in each case. I do not know how to find the probability of choosing n particular choices( all less than or equal to 9) out of m choices remembering the fact that 9 should appear in atleast one of the 7 draws

Chen Guo
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1 Answers1

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For a with replacement selection each draw is independent. Consider the $k^{th }$ draw. Any ticket numbered less than $9$ can be chosen, the probability of chosing such a ticket will be $\frac9{15}$, as any ticket has the same probability, $\frac1{15}$, of being chosen. Hence the probability that in all the $7$ draws the tickets chosen are $\le9$, will be $$\left(\frac9{15}\right)^7$$ Similarly, the probability that in all the $7$ draws the tickets chosen are $\le8$, will be $$\left(\frac8{15}\right)^7$$ But see that in the former case the maximum of the numbers drawn will he $\le9$, while in the latter case the maximum of the numbers drawn will be $\le8$. So the probability that the maximum of the numbers drawn is exactly $9$ will be $$\left(\frac9{15}\right)^7- \left(\frac8{15}\right)^7$$

QED
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  • Does this guarantee 9 in each draw? Here we can pick 10 also...So for 9 we need 7C1 i guess – Chen Guo Jan 14 '18 at 05:59
  • @Abhishanka Saha That does not explain correctly...i think possible outcomes are 9^7 - 8^7 – Chen Guo Jan 14 '18 at 06:27
  • See I just found out the probability that each draw is $\le9$. I didn't complete the problem, so that the OP can themselves do it – QED Jan 14 '18 at 07:05
  • Anyways I am editing the answer, so that I don't get anymore negative markings – QED Jan 14 '18 at 07:05