Gene has less than 12 nickels,dimes,and quarters in his pocket.The probability of pulling out a nickel or a quarter is 3/4.The probabilty of pulling out a dime is 1/4.How many coins does Gene have in his pocket?How many of each does he have?Please explain

For your first question, the first sentence answers it, "12 nickels,dimes,and quarters in his pocket."

For the dime, 1/4 of 12 = 3

As I indicated before, there is not enough data to tell how many of the remaining 9 coins are nickels or quarters. Do you have more information that you have not included, for instance, the total value of the coins?

To solve this problem, we'll use algebra and logical reasoning.

Let's assume Gene has x nickels, y dimes, and z quarters in his pocket. Since the problem states that he has less than 12 coins in total, we can write the following inequality:

x + y + z < 12 (Equation 1)

Now, the probability of pulling out a nickel or a quarter is 3/4, which means there is a 3/4 chance of pulling out either a nickel or a quarter. Since there are only three types of coins in consideration (nickels, dimes, and quarters), the probability of pulling out a dime is therefore 1 - 3/4 = 1/4.

Also, we know that the total probability of an event happening is equal to the sum of the probabilities of its individual outcomes. In other words, the probability of pulling out a nickel, plus the probability of pulling out a dime, plus the probability of pulling out a quarter, should add up to 1.

The probability of pulling out a nickel is (x)/(x+y+z), where x represents the number of nickels Gene has. Similarly, the probability of pulling out a dime is (y)/(x+y+z), and the probability of pulling out a quarter is (z)/(x+y+z).

So, we can write the equation:

(x)/(x+y+z) + (y)/(x+y+z) + (z)/(x+y+z) = 1/4 (Equation 2)

Now, let's solve Equations 1 and 2 simultaneously to find the values of x, y, and z.

Since Equation 1 states that the sum of the numbers of nickels, dimes, and quarters is less than 12, we need to consider the possible combinations.

One possible combination that satisfies Equation 1 is when x = 0, y = 3, and z = 8. Let's substitute these values in Equation 2:

(0)/(0+3+8) + (3)/(0+3+8) + (8)/(0+3+8) = 1/4

0 + 3/11 + 8/11 = 1/4

11/11 = 1/4

This combination satisfies Equation 2.

So, Gene has 0 nickels, 3 dimes, and 8 quarters, which adds up to a total of 11 coins.