Question 1183647: Please help me solve this. John forgot the last three digits of his friend's telephone number. He only remembers that the digits are not the same and are not even. What is the probability he dials the correct telephone number on the first try?
Found 3 solutions by ikleyn, robertb, greenestamps:
Answer by ikleyn(52814)   (Show Source):
Answer by robertb(5830)  (Show Source):
You can put this solution on YOUR website! Dialing phone numbers works on the basis of the correct order of the digits.
Hence, the phone number XXXXXXX135 would be different from the number XXXXXXX315, or the number XXXXXXX531.
In other words, the probability that John dials the correct phone number on the first try is  ~ 0.0167, to 4 d.p.
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
I will disagree with both of the answers provided by the other tutors.
Certainly the one that treats the problem as combinations instead of permutations is incorrect, since the order of the digits is important in a phone number.
And I will disagree with the other response also, due to a different interpretation of the problem.
The statement of the problem says the three digits "are not the same". Grammatically, that is not the same as saying that they are "all different". Saying the three digits are "not the same" allows the possibility that two of them are the same but the third is different.
So the total number of possible sequences of the last three digits, given that they are all odd, is 5*5*5=125; and of those, only the five sequences 111, 333, 555, 777, and 999 are not allowed.
So the number of allowable sequences of the last three digits is 125-5=120; and since only one of those sequences is the correct one, the probability of dialing the correct phone number is 1/120.
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