Coin box Question: Assume that there are 25 pennies, 10 nickels, 10 dimes, and 5 quarters in a box. after shuffeling the box, you pick a coin at random and note down its face value. let x represent the face value of the coin in cents. then construct the probability distribution of x. The question I need to find is...Find P(x is at most 12) ? I am not sure what to do or how to do it. i would appreciate the help. Thanks,
To find the probability distribution of x, we need to calculate the probability of selecting each possible face value from the given set of coins.
Here's how you can construct the probability distribution:
Step 1: Count the total number of coins in the box:
Total coins = 25 (pennies) + 10 (nickels) + 10 (dimes) + 5 (quarters) = 50 coins.
Step 2: Find the probability of selecting each coin:
- Probability of selecting a penny = Number of pennies / Total coins = 25/50 = 1/2 = 0.5 (or 50%).
- Probability of selecting a nickel = Number of nickels / Total coins = 10/50 = 1/5 = 0.2 (or 20%).
- Probability of selecting a dime = Number of dimes / Total coins = 10/50 = 1/5 = 0.2 (or 20%).
- Probability of selecting a quarter = Number of quarters / Total coins = 5/50 = 1/10 = 0.1 (or 10%).
Now that we have the probability distribution, let's find P(x is at most 12):
Step 3: Identify the face values that are at most 12 cents:
The face values that are at most 12 cents are pennies (1 cent), nickels (5 cents), and dimes (10 cents).
Step 4: Calculate the probability of selecting a coin with a face value at most 12 cents:
P(x is at most 12) = P(1 cent) + P(5 cents) + P(10 cents)
= Probability of selecting a penny + Probability of selecting a nickel + Probability of selecting a dime
= 0.5 + 0.2 + 0.2 (since there are no quarters with face value less than or equal to 12 cents)
= 0.9 (or 90%).
Therefore, the probability of selecting a coin with a face value at most 12 cents is 0.9 (or 90%).