MHB Stuck on a probability question - scrambled phone number

AI Thread Summary
The discussion revolves around calculating the probability of correctly guessing a scrambled phone number, specifically 123-4567, where the first three digits and the last four digits are mixed. The user has determined the probability of selecting all numbers correctly as 1/3!4! and the probability of getting the first three correct as 1/3!. The challenge lies in calculating the probability of having the first three digits correct while also having at least one of the last four digits correct. The user initially miscalculated this probability but later found the correct answer to be 15/3!4! by listing possible combinations. The thread seeks clarification on how to derive this answer using probability principles.
fbab
Messages
1
Reaction score
0
The question goes like this:
Suppose a phone number is 123-4567, and that the first three numbers and last 4 numbers are scrambled.

First parts I figured out:
Probability of choosing every number correctly: 1/3!4!
probability of choosing of choosing the first 3 correctly:1/3!

Where I got stuck on, the probability of having the first three correct, and at least 1 of the numbers in the last 4 correct. My initial reaction was that it would be 1/3!* the probability of not getting any correct in the last 4. I just do not know how to calculate this, I initially thought it would be (3/4)*(2/3)*(1/2) which would be 1/4 which would have given me a final answer of 18/3!4!.

Looking at the answer in the back of the book showed me the answer was 15/3!4!. I now know this is the answer by writing down all the possible options, but would love to know how to come by this answer with properties of probability.

Any help would be greatly appreciated!
 
Mathematics news on Phys.org
Re: Stuck on a prob question

fbab said:
The question goes like this:
Suppose a phone number is 123-4567, and that the first three numbers and last 4 numbers are scrambled.

First parts I figured out:
Probability of choosing every number correctly: 1/3!4!
probability of choosing of choosing the first 3 correctly:1/3!

Where I got stuck on, the probability of having the first three correct, and at least 1 of the numbers in the last 4 correct. My initial reaction was that it would be 1/3!* the probability of not getting any correct in the last 4. I just do not know how to calculate this, I initially thought it would be (3/4)*(2/3)*(1/2) which would be 1/4 which would have given me a final answer of 18/3!4!.

Looking at the answer in the back of the book showed me the answer was 15/3!4!. I now know this is the answer by writing down all the possible options, but would love to know how to come by this answer with properties of probability.

Any help would be greatly appreciated!
Hello fbab,
Welcome to MHB!
What is the probability that having
1xx-xxxx
x2x-xxxx
xx3-xxxx

Regards
$$|\rangle$$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top