Question

In a lottery, a 4 digit number from 0000 to 9999 is randomly selected. Find the probability that the number selected begins with a 3 and ends in a 2 or 0.

Answer 1

The probability that a randomly selected 4-digit number begins with a 3 and ends with a 2 or 0 is 0.0200 or 2%.

Step-by-step explanation:

To find the probability that a randomly selected 4-digit number begins with a 3 and ends in a 2 or 0, we can use the rule of product for independent events. Since the first digit must be 3, there is one choice for the first digit. For the second and third digits, they can be any number from 0 to 9, so we have 10 choices for each. The last digit can be either 2 or 0, which provides 2 possibilities.

Thus, the total number of favorable outcomes is 1 (for the first digit) × 10 (for the second digit) × 10 (for the third digit) × 2 (for the last digit) = 200 favorable outcomes. The total number of possible 4-digit numbers is 10000 (ranging from 0000 to 9999).

So the probability is calculated as the number of favorable outcomes divided by the number of total possible outcomes: Probability = 200 / 10000 = 0.02. After rounding to four decimal places, as instructed, the probability is 0.0200 or 2%.

Answer 2

Since the choices is from 0000 to 9999, therefore there are a total of 10,000 possibilities.

The possibilities that the number begins with a 3 and ends in a 2 or 0 is:

possibilities = 1 * 10 * 10 * 1 + 1 * 10 * 10 * 1 = 200

Therefore the probability is:

P = 200 / 10,000 = 0.02 = 2%

So there is a 2% probability.