A lock is opened when we enter a two digit key. The first digit of the key is an alphabet and the second digit is an even number from 0 through 20. The probability of choosing the key through the above $condition is given as \frac{1}{A} . What is the value of 'A'?$
260

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Solution

The correct option is A 260
The key of the lock is composed of a 2 digit number.

Given, first digit is an alphabet.
There are 26 alphabets.

So, first digit can be chosen from A through Z in 26 different ways.

The second digit of the key is an even number from 0 through 20. There are 10 even numbers between 0 and 20, starting from 0.

So, the second digit can be chosen in 10 ways.

∴ Total number of outcome will be
$26 \times 10 = 260$

Only one time, we can get the key perfectly correct.

So, the probability of to getting the correct key is
$P = \frac{1}{260} = \frac{1}{A}$

Therefore, value of 'A' = 260

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A lock is opened when we enter a two digit key. The first digit of the key is an alphabet and the second digit is an even number from 0 through 20. The probability of choosing the key through the above condition is given as 1A. What is the value of 'A'?260

A lock is opened when we enter a two digit key. The first digit of the key is an alphabet and the second digit is an even number from 0 through 20. The probability of choosing the key through the above $condition is given as \frac{1}{A} . What is the value of 'A'?$
260