Repeating numbers.

[Guest]

Hello there.

am wanting to find the probability of that out of 26 random numbers, 22 end with the same digit.

As in I have 26 numbers (each a 5 digit code) 22 of which are ending with a '7' but the other four not a seven are random.

I'm not sure how to calculate this.

Thanks

[shyamjayakannan]

total possibilities = (number of possible 5 digit numbers)[tex]^{26}[/tex] = [tex]\left(9\times10^4\right)^{26}[/tex]
Now, for 22 to have the same last digit: (choose 22 of the 26) x (choose the common last digit) x (choose the remaining digits for them) x (number of possible remaining 4 numbers)
[tex]={}^{26}C_{22}\times10\times\left(9\times10^3\right)^{22}\times\left(9\times10^4\right)^4[/tex]

So, probability = [tex]\frac{^{26}C_{22}\times10\times\left(9\times10^3\right)^{22}\times\left(9\times10^4\right)^4}{\left(9\times10^4\right)^{26}}=\boxed{\frac{^{26}C_{22}}{10^{22}}}[/tex]