Questions: A box contains 6 green marbles and 14 white marbles. If the first marble chosen was a white marble, what is the probability of choosing, without replacement, another white marble? Express your answer as a fraction or a decimal number rounded to four decimal places.

Solution Steps

We have a box containing 6 green marbles and 14 white marbles, making a total of \( N = 20 \) marbles. After drawing one white marble, we need to find the probability of drawing another white marble without replacement.

After drawing one white marble, the counts of the marbles are updated as follows:

• Total marbles left: \( N = 20 - 1 = 19 \)
• White marbles left: \( K = 14 - 1 = 13 \)
• Green marbles remain unchanged: 6

We want to calculate the probability of drawing \( k = 1 \) white marble from the remaining \( n = 1 \) draw. The hypergeometric probability formula is given by:

\[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \]

Substituting the values:

• \( K = 13 \) (remaining white marbles)
• \( k = 1 \) (white marbles drawn)
• \( N = 19 \) (total remaining marbles)
• \( n = 1 \) (total marbles drawn)

We compute the probability as follows:

\[ P(X = 1) = \frac{\binom{13}{1} \binom{6}{0}}{\binom{19}{1}} = \frac{13 \cdot 1}{19} = \frac{13}{19} \approx 0.6842 \]

Final Answer