jordan has 2 quarters, 1 nickel, and 3 dimes. if he took 1 out and then a second one without replacing the first coin, what is the probability that both coins were dimes?
3/6 * 2/5= 1/2 * 2/5= 2/10 = 1/5
well first you find how many coins there are in all. In this problem the entire amount is 6 coins,
3 of the 6 coins are dimes,
if you pull out one dime there are now 2 dimes out of 5 coins ( because you took one away)
so the answer would be two out of five chances.
Cycv
To find the probability of drawing two dimes consecutively without replacement, we first need to determine the total number of coins available and the total number of possible outcomes.
In this scenario, Jordan has 2 quarters, 1 nickel, and 3 dimes. So, the total number of coins available for selection is 2 + 1 + 3 = 6.
Now, let's calculate the probability:
Step 1: Calculate the probability of the first coin being a dime.
Since Jordan has a total of 6 coins and 3 of them are dimes, the probability of selecting a dime as the first coin is 3/6 or 1/2.
Step 2: Calculate the probability of the second coin being a dime without replacing the first coin.
After removing the first coin, Jordan is left with a total of 5 coins, out of which 2 are dimes. Therefore, the probability of selecting a dime as the second coin is 2/5.
Step 3: Calculate the probability of both coins being dimes consecutively.
To find the probability of multiple independent events occurring together, we multiply the individual probabilities. So, the probability of both coins being dimes is (1/2) * (2/5) = 1/5.
Therefore, the probability that both coins drawn consecutively without replacement are dimes is 1/5.