A jar contains 2 pennies, 6 nickels and 4 dimes. A child selects 2 coins at random without replacement from the jar. Let X represent the amount in cents of the selected coins.
a. Find the probability X = 11.
b. Find the expected value of X
A jar contains 7 pennies, 7 nickels and 4 dimes. A child selects 2 coins at random without replacement from the jar. Let X represent the amount in cents of the selected coins.
a. To find the probability that X = 11, we need to find the number of ways we can select 2 coins that add up to 11 cents and divide it by the total number of possible selections.
There are two cases where the selected coins add up to 11 cents:
1. Selecting one nickel and one dime: 6 ways to select a nickel and 4 ways to select a dime.
2. Selecting one penny, one nickel, and one dime: 2 ways to select a penny, 6 ways to select a nickel, and 4 ways to select a dime.
So, the total number of ways we can select 2 coins that add up to 11 cents is 6 + 2 = 8.
The total number of possible selections is given by the combination formula:
C(12, 2) = 12! / (2! * (12-2)!) = 66
Therefore, the probability that X = 11 is 8/66, which simplifies to 4/33.
b. To find the expected value of X, we need to calculate the average value of X over all possible selections.
Let's assign a value to each coin as follows:
- Penny: 1 cent
- Nickel: 5 cents
- Dime: 10 cents
The possible values of X are:
- If we select two pennies: X = 1 + 1 = 2 cents
- If we select one penny and one nickel: X = 1 + 5 = 6 cents
- If we select one penny and one dime: X = 1 + 10 = 11 cents
- If we select two nickels: X = 5 + 5 = 10 cents
- If we select one nickel and one dime: X = 5 + 10 = 15 cents
- If we select two dimes: X = 10 + 10 = 20 cents
We can calculate the probability for each of these scenarios as follows:
- Probability of selecting two pennies: (2/12) * (1/11) = 1/66
- Probability of selecting one penny and one nickel: (2/12) * (6/11) + (6/12) * (2/11) = 12/66 = 2/11
- Probability of selecting one penny and one dime: (2/12) * (4/11) + (4/12) * (2/11) = 8/66 = 4/33
- Probability of selecting two nickels: (6/12) * (5/11) = 5/22
- Probability of selecting one nickel and one dime: (6/12) * (4/11) + (4/12) * (6/11) = 24/66 = 4/11
- Probability of selecting two dimes: (4/12) * (3/11) = 1/11
Now, we can calculate the expected value of X:
Expected value = (2 * (1/66) + 6 * (2/11) + 11 * (4/33) + 10 * (5/22) + 15 * (4/11) + 20 * (1/11))
Simplifying this expression gives the expected value of X as 7 cents.
To find the probability of X = 11, we need to determine the number of favorable outcomes and the total number of possible outcomes.
Step 1: Determine the number of favorable outcomes
For X = 11, we can have the following combinations of coins:
- Penny (1 cent) + Dime (10 cents)
- Nickel (5 cents) + Nickel (5 cents)
Step 2: Calculate the number of ways to select 2 coins out of the total number of coins
The total number of coins in the jar is 2 pennies + 6 nickels + 4 dimes = 12 coins. To select 2 coins out of these 12, we use binomial coefficients or combinations. The number of ways to select 2 coins from 12 is given by "12 choose 2," which can be calculated as:
12C2 = 12! / (2! * (12-2)!)
Step 3: Find the probability
The probability of X = 11 can be calculated as the number of favorable outcomes divided by the number of total outcomes:
P(X = 11) = Number of favorable outcomes / Number of total outcomes
P(X = 11) = (number of combinations for Penny + Dime + number of combinations for Nickel + Nickel) / 12C2
Now let's calculate the values.
For Penny + Dime:
The number of combinations for selecting a Penny = 2C1 = 2 (since there are 2 pennies in the jar)
The number of combinations for selecting a Dime = 4C1 = 4 (since there are 4 dimes in the jar)
For Nickel + Nickel:
The number of combinations for selecting 2 nickels = 6C2 = 15 (since there are 6 nickels in the jar)
Number of favorable outcomes = 2 (Penny + Dime) + 15 (Nickel + Nickel) = 17
Total number of outcomes = 12C2 = 12! / (2! * (12-2)!) = 66
P(X = 11) = 17 / 66
P(X = 11) ≈ 0.258 or 25.8%
For part b, to calculate the expected value of X, we need to multiply each possible value of X by its corresponding probability and sum them up.
We have the following possible values for X:
- When X = 1 (Penny + Penny): Value = 1 cent, Probability = (2C2 / 12C2) = 1/66
- When X = 5 (Nickel + Nickel): Value = 10 cents, Probability = (6C2 / 12C2) = 15/66
- When X = 11 (Penny + Dime, Nickel + Nickel): Value = 11 cents, Probability = (17 / 66) calculated in part a
Now, let's calculate the expected value:
Expected value of X = (1 * (1/66)) + (10 * (15/66)) + (11 * (17/66))
Expected value of X = (1/66) + (150/66) + (187/66)
Expected value of X = 5.98 or approximately 6 cents.
If x = 11 for two coins, they must be a penny and a dime.
P(dime) = 4/12
P(penny) = 2/11 because one coin already removed from jar.
X = 4/12 *2/11 = ?