Question

There are 4 red, 3 blue and 3 yellow marbles in an urn. If three marbles are drawn simultaneously, then the probability that the number of yellow marbles will be less than 2 is equal to

• A. \(\frac{97}{120}\)
• B. \(\frac{49}{60}\)
• C. \(\frac{47}{60}\)
• D. \(\frac{59}{60}\)
• E. \(\frac{39}{60}\)

Answer 1

The Correct Option is B

We are given:
- 4 red marbles,
- 3 blue marbles,
- 3 yellow marbles.

Three marbles are drawn simultaneously, and we need to find the probability that the number of yellow marbles drawn will be less than 2. This means we are interested in cases where either 0 or 1 yellow marble is drawn.

Step 1: Total number of ways to draw 3 marbles
The total number of ways to draw 3 marbles from a set of 10 (4 red + 3 blue + 3 yellow) marbles is given by the combination:
 

\[\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120.\]

Step 2: Number of favorable outcomes
We need to calculate the number of ways to draw 0 or 1 yellow marble.

Case 1: Drawing 0 yellow marbles
If 0 yellow marbles are drawn, we must select all 3 marbles from the red and blue marbles. There are 7 marbles (4 red and 3 blue), and the number of ways to choose 3 marbles from these 7 is:
 

\[\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35.\]

Case 2: Drawing 1 yellow marble
If 1 yellow marble is drawn, we must select 2 marbles from the 7 red and blue marbles. The number of ways to choose 2 marbles from these 7 is:
 

\[\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21.\]

Then, there are 3 ways to choose 1 yellow marble from the 3 yellow marbles:
 

\[\binom{3}{1} = 3.\]

So, the total number of favorable outcomes for this case is:
 

\[21 \times 3 = 63.\]

Step 3: Total favorable outcomes
The total number of favorable outcomes is the sum of the outcomes from Case 1 and Case 2:
 

\[35 + 63 = 98.\]

Step 4: Calculate the probability
The probability that the number of yellow marbles drawn is less than 2 is the ratio of favorable outcomes to total outcomes:
 

\[\frac{98}{120} = \frac{49}{60}.\]

So, the correct option is (B) : \(\frac{49}{60}\)

Answer 2

There are 4 red, 3 blue, and 3 yellow marbles in an urn. A total of 4 + 3 + 3 = 10 marbles.

We want to find the probability that the number of yellow marbles drawn is less than 2, i.e., either 0 or 1 yellow marbles are drawn when we draw 3 marbles simultaneously.

The total number of ways to choose 3 marbles out of 10 is \(\binom{10}{3} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120\).

Case 1: 0 yellow marbles are drawn. This means we choose 3 marbles from the 4 red and 3 blue marbles (7 marbles total). The number of ways to do this is \(\binom{7}{3} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 7 \times 5 = 35\).

Case 2: 1 yellow marble is drawn. This means we choose 1 yellow marble from the 3 yellow marbles and 2 marbles from the remaining 7 (4 red and 3 blue). The number of ways to do this is \(\binom{3}{1} \times \binom{7}{2} = 3 \times \frac{7!}{2!5!} = 3 \times \frac{7 \times 6}{2 \times 1} = 3 \times 21 = 63\).

The number of ways to have less than 2 yellow marbles is the sum of the number of ways in Case 1 and Case 2: 35 + 63 = 98.

The probability that the number of yellow marbles will be less than 2 is:

\(P(\text{less than 2 yellow}) = \frac{\text{number of ways to have less than 2 yellow marbles}}{\text{total number of ways to choose 3 marbles}} = \frac{98}{120} = \frac{49}{60}\)