A few days ago
Human No.2543324757

# Math.help!you dont have to answer the question but i need help?

Employee identification codes at a company contain 2 letters followed by 2 numbers.all codes are equally likely.

1.find the number of possible identification codes.

2.find the probability of beign assigned the code MT49

3.find the probability that an ID code of the company does not contain the letter A as the second letter of the code.

4.find the probablility that an ID code of the company does not contain the letter 2.

you dont have to give me the answers i just need help.but if you know the answers plese post em 😛

i could not see the board when they explain it on class i need glasses and im getting some

A few days ago
Anand S

The total number of combinations is found by multiplying together the number of possibilities for each position.

Say the code were 4 numerical digits. That means each position can only have 10 possible entries: 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. So there are 10x10x10x10 = 10,000 different codes possible.

In your case, there are 26 choices for each letter slot and 10 choices for each number slot.

For the other problems, all that changes is the number of possibilities for a particular slot. In the case of number 3, if you take away A as a possible letter, that leaves only 25 possibilities for the second slot.

The probability of something happening is calculated by taking all the *desired* outcomes and dividing by all the *possible* outcomes (i.e., dividing by the answer to part 1).

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A few days ago
stylestj
OK… here ya go!

1. There are 26 letters in the alphabet

The numbers are 0 – 9

so the formula to figure out the number of possible codes is 26 X 26 X 10 X 10

2. Solve the equation and use the info to answer the questions

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