Question

A box contains 3 identical green balls and 7 identical blue balls. Two balls are randomly drawn without replacement from the box. The probability of drawing 1 green and 1 blue ball is

• A. $\frac{3P1 \times 7P1}{10P2}$
• B. $\frac{10P3 \times 10P7}{10P2}$
• C. $\frac{10C3 \times 10C7}{10C2}$
• D. $\frac{3C1 \times 7C1}{10C2}$

Hint:

Combinations (\(\binom{n}{k}\)) are used when the order of selection does not matter, which is typically the case in such probability problems where objects are identical within groups.

Answer 1

The Correct Option is D

To find the probability of drawing one green and one blue ball from a box containing 3 green and 7 blue balls, we consider the problem in terms of combinations because the balls are identical within their colors. 
Step 1: Calculate the number of ways to choose 1 green ball from 3. This can be done using the combination formula \(\binom{n}{k}\): \[ \binom{3}{1} = 3 \] Step 2: Calculate the number of ways to choose 1 blue ball from 7: \[ \binom{7}{1} = 7 \] Step 3: Calculate the total number of ways to draw any two balls from the 10 balls (3 green + 7 blue): \[ \binom{10}{2} = 45 \] Step 4: Compute the probability of drawing one green and one blue ball: \[ \frac{\binom{3}{1} \times \binom{7}{1}}{\binom{10}{2}} = \frac{3 \times 7}{45} = \frac{21}{45} = \frac{7}{15} \] Therefore, the probability of drawing one green and one blue ball is \(\frac{7}{15}\), and the correct option based on this analysis is: \(\frac{3C1 \times 7C1}{10C2}\)