What is the probability that a phone number generated using the digits
1, 2, 2, 4, 5, 5, 6, and 2 is the number 654-5222?
+118723 What is the probability that a phone number generated using the digits
1, 2, 2, 4, 5, 5, 6, and 2 is the number 654-5222?
mmm
\(\frac{1}{8}\times \frac{2}{7}\times \frac{1}{6}\times \frac{1}{5}\times \frac{3}{4}\times \frac{2}{3}\times \frac{1}{2}\\ =\frac{1*2*1*1*3*2*1}{8!}\\ =\frac{12}{40320}\\ =\frac{1}{336}\\\)
OR
The number of permutations for choosing 7 digits from these digits is
\(\frac{8P7}{3P3*2P2}=\frac{8!}{3!2!}=\frac{40320}{6*2}=336\)
We want the probability of choosing just one of these permutations so the answer is \(\frac{1}{336}\)
+37170 I come up with 1 out of 40,320 chance.
nPr =
8P7 = 40320
1st digit 8 choices
2nd 7
3rd 6
4th 5
5th 4
6th 3
7th 2 8x7x6x5x4x3x2 = 40320 possible combinations of which ONE will be 654 5222
+118723 What is the probability that a phone number generated using the digits
1, 2, 2, 4, 5, 5, 6, and 2 is the number 654-5222?
mmm
\(\frac{1}{8}\times \frac{2}{7}\times \frac{1}{6}\times \frac{1}{5}\times \frac{3}{4}\times \frac{2}{3}\times \frac{1}{2}\\ =\frac{1*2*1*1*3*2*1}{8!}\\ =\frac{12}{40320}\\ =\frac{1}{336}\\\)
OR
The number of permutations for choosing 7 digits from these digits is
\(\frac{8P7}{3P3*2P2}=\frac{8!}{3!2!}=\frac{40320}{6*2}=336\)
We want the probability of choosing just one of these permutations so the answer is \(\frac{1}{336}\)
