Question

A bag contains 5 yellow, 3 green, 2 blue and 7 white balls. If 4 balls are chosen at random, then the probability that none of them are white is

• A. \(\frac{3}{37}\)
• B. \(\frac{7}{34}\)
• C. \(\frac{5}{34}\)
• D. \(\frac{5}{37}\)
• E. \(\frac{3}{34}\)

Answer 1

Answer: (E)

Given: A bag contains 5 yellow, 3 green, 2 blue, and 7 white balls. A total of 5 + 3 + 2 + 7 = 17 balls are present.

Step 1: Total ways to choose 4 balls from 17 balls:

\[ \text{Total ways} = \binom{17}{4} = \frac{17 \times 16 \times 15 \times 14}{4 \times 3 \times 2 \times 1} = 2380 \]

Step 2: Ways to choose 4 balls such that none are white (i.e., choosing from 10 non-white balls):

\[ \text{Ways to choose non-white balls} = \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \]

Step 3: Calculate the probability:

\[ \text{Probability} = \frac{\text{Ways to choose non-white balls}}{\text{Total ways}} = \frac{210}{2380} = \frac{3}{34} \]

Final Answer:

\[ \frac{3}{34} \]