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Suppose 5 different integers are randomly chosen from between 20 and 69, inclusive. What is the probability that they each have a different tens digit?

 Jan 20, 2023
 #1
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There are 50 integers to choose from between 20 and 69, inclusive. To have a different tens digit, the integers must be chosen from the set {20, 21, 22, ... , 29, 30, 31, ... , 69}.  We can select the first integer in 50 ways. The second integer must be chosen from 49 remaining integers, the third integer from 48 remaining integers and so on.

 

So the probability of selecting 5 different integers with different tens digit is (50*49*48*47*46)/(50^5) = 317814/390625.

 Jan 20, 2023
 #2
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That's not right but thanks

Could someone else help plase?

dolphinia  Jan 22, 2023
 #3
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FIRSTLY Delphinia,  do not down vote answers just becasue you think it is wrong.

If a person makes a serious attempt at answering then you should thank them and not punish them in ANY way.

I answered you before I realiszed you had done this, otherwise I probably would not have answered you.

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Suppose 5 different integers are randomly chosen from between 20 and 69, inclusive. What is the probability that they each have a different tens digit?

 

there are  69-20+1= 50 numbers altogether

There are 5 different 10 digits,   2,3,4,5,6

 

Choose a number any number  prob = 1

 

now there are  49 numbers left and 40 of them have a tens unit not yet chose, 

P of the next one being a good choice is  40/49

and so on:

 

\(Prob = 1 * \frac{40}{49}*\frac{30}{48}*\frac{20}{47}*\frac{10}{46}\)

 

You can finish it. 

 Jan 24, 2023

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