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.
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?
1 answer