I have a quarter, three dimes, two nickles in a bag. I pick a coin from the bag. Then, without replacing it, pick another coin. What is the probability of picking a quarter, then a dime?

Answer 1

So this problem is really only solvable if we assume that each coin in the bag has an equal likelihood of being picked on each individual draw -- an assumption that may not necessarily be true in actuality because the size of each coin may affect how likely it is to be picked. (A quarter might be more likely than a dime to be picked for example, because it is so much larger.)

Barring that, assuming the size of the coin does not affect the likelihood, the probability calculation we're confronted with is actually quite straightforward. We are basically finding the probability that we pick a quarter, then, on the next draw, a dime. The wording of the question implies that it does occur in that order, so drawing a dime, and and then a quarter, does not count.

We can use the multiplication rule of probability to solve this. The multiplication rule goes as follows :

" Rule of Multiplication The probability that Events A and B both occur is equal to the probability that Event A occurs times the probability that Event B occurs, given that A has occurred.

P(A ∩ B) = P(A) P(B|A) "

Source: https://stattrek.com/probability/probability-rules.aspx#:~:text=Rule%20of%20Multiplication%20The%20probability,given%20that%20A%20has%20occurred.

In layman's terms: The probability of two things happening,is the the probability of the first thing happening times the probability of the second thing happening given the first thing happened.

So for this example, the first thing that happens is we draw a quarter. To get the chance of drawing a quarter we must know the total number of coins in the bag -- Remember, a simple probability like this where each coin is equally likely is calculated by doing # successes / # total.

There is 1 quarter in the bag. (our success) and the total number of coins is :

1 quarter + 3 dimes + 2 nickels = 6 total coins.

Therefore, the probability of drawing a quarter is 1 / 6 = 1/6.

Now, for our next draw there is now 1 less coin in the bag, one less quarter, since we didn't replace coins after the first draw.

3 dimes + 2 nickels = 5 total remaining coins.

(Remember we have to find the chance of the second thing happening given the first thing already happened.)

For this draw, we need to draw a dime. The probablity of drawing a dime now is 3 dimes / 5 total coins = 3/5

So to get our total probablity we take our first draw probability, 1/6, and multiply it by our 2nd draw probability, 3/5.

1/6 * 3/5 = 3/30 = 1/10 or 0.1.

Hope that helps!

Answer 2

There are a total of 6 coins (1 quarter, 3 dimes, and 2 nickels) in the bag.

I will randomly select 2 coins from the bag.

Q = picking a quarter (1/6)

D = picking a dime (3/6 = 1/2)

N = picking a nickel (2/6 = 1/3)

P(Q ∩ D) = P(Q)*P(D|Q) = (1/6)*(3/5) = 3/30 = 1/10 = 0.10

The probability of selecting a quarter first and then a dime without replacement is 0.10.

Answer 3

The probability of getting a quarter is number of quarters/total number of coins = 1/(1+3+2) = 1/6

The probability of getting a dime after that is number of dimes/total number of coins. Since there is no replacement, the denominator reduces by 1 = 3/(2+3) = 3/5

To get the probability of both happening, multiply them

(1/6)(3/5) = 3/10 = 1/10