Questions: - Suppose you like to keep a jar of change on your desk. Currently, the jar contains the following: 17 Pennies 10 Dimes 19 Nickels 21 Quarters Copy Data What is the probability that you reach into the jar and randomly grab a nickel and then, without replacement another nickel? Express your answer as a fraction or a decimal number rounded to four decimal phices.

Solution Steps

The total number of coins in the jar is calculated as follows: \[ N = 17 \text{ (Pennies)} + 10 \text{ (Dimes)} + 19 \text{ (Nickels)} + 21 \text{ (Quarters)} = 67 \] The number of nickels is given as: \[ K = 19 \]

The probability of drawing the first nickel is given by: \[ P(\text{First Nickel}) = \frac{K}{N} = \frac{19}{67} \approx 0.2836 \]

After drawing the first nickel, there are now 18 nickels left and a total of 66 coins remaining. The probability of drawing the second nickel is: \[ P(\text{Second Nickel}) = \frac{K - 1}{N - 1} = \frac{18}{66} \approx 0.2727 \]

The combined probability of drawing two nickels in succession without replacement is: \[ P(X = 2) = P(\text{First Nickel}) \times P(\text{Second Nickel}) = \frac{19}{67} \times \frac{18}{66} \approx 0.0773 \]

Using the hypergeometric distribution, the probability of drawing exactly 2 nickels from a total of 67 coins, where 19 are nickels, is calculated as: \[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} = \frac{\binom{19}{2} \binom{48}{0}}{\binom{67}{2}} \approx 0.0773 \]

Final Answer