A bag contains 7 nickels, 2 dimes, and 8 quarters. If you draw 3 coins at random from the bag, without replacement, what is the probability that you will get a nickel (N), a quarter (Q), and a nickel (N), in that order?
There are a total of 7 + 2 + 8 = <<7+2+8=17>>17 coins in the bag.
To calculate the probability of drawing a nickel, a quarter, and a nickel in that order, we need to consider the number of ways we can do so and divide it by the total number of possible outcomes.
Since we are drawing without replacement, the first nickel can be selected from the 7 nickels in the bag.
After the first nickel is drawn, there are 16 coins left in the bag (since the first nickel is not replaced). Out of these 16 coins, 8 are quarters.
Finally, after drawing a quarter, there are 15 coins left in the bag (since both the nickel and the quarter are not replaced). Out of these 15 coins, 6 are nickels.
Therefore, there are 7 * 8 * 6 = <<7*8*6=336>>336 ways to draw a nickel, a quarter, and a nickel in that order.
The total number of ways to draw 3 coins from the bag without replacement is given by the combination formula C(17, 3) = 17! / (3! * (17-3)!) = 17! / (3! * 14!) = (17 * 16 * 15) / (3 * 2 * 1) = 2720.
Therefore, the probability of drawing a nickel, a quarter, and a nickel in that order is 336 / 2720 = <<336/2720=0.12352941176>>0.1235 (rounded to 4 decimal places).
So the probability is approximately 0.1235.
To find the probability of drawing a nickel, then a quarter, and then another nickel in that order, we can use the concept of Conditional Probability.
Step 1: Find the probability of drawing a nickel (N) as the first coin.
There are a total of 7 nickels out of 17 coins (7 nickels, 2 dimes, and 8 quarters) in the bag.
So, the probability of drawing a nickel as the first coin = Number of nickels / Total number of coins = 7/17.
Step 2: Find the probability of drawing a quarter (Q) as the second coin, given that a nickel has already been drawn.
After drawing the first nickel, there are 6 nickels, 2 dimes, and 8 quarters remaining in the bag, totaling 16 coins.
So, the probability of drawing a quarter as the second coin = Number of quarters left / Total number of coins left = 8/16 = 1/2.
Step 3: Find the probability of drawing another nickel (N) as the third coin, given that a nickel and a quarter have already been drawn.
After drawing the first nickel and the quarter, there are 5 nickels, 2 dimes, and 7 quarters remaining in the bag, totaling 14 coins.
So, the probability of drawing another nickel as the third coin = Number of nickels left / Total number of coins left = 5/14.
Step 4: Multiply the probabilities from each step to find the overall probability of drawing a nickel, then a quarter, and then another nickel in that order.
Probability = (Probability of drawing a nickel as the first coin) × (Probability of drawing a quarter as the second coin) × (Probability of drawing another nickel as the third coin)
= (7/17) × (1/2) × (5/14)
Calculating this expression, we get:
Probability = 0.0735
So, the probability of drawing a nickel, then a quarter, and then another nickel in that order is 0.0735 or 7.35%.