A bag contains 7 blue balls and 5 yellow balls , if two balls are selected at random, what is the probability that none is yellow ?

Options

A. `(5)/(35)`
B. `(5)/(22)`
C. `(7)/(22)`
D. `(7)/(33)`

Explanation

To solve the problem step by step, we will calculate the probability that none of the selected balls are yellow when choosing 2 balls from a bag containing 7 blue balls and 5 yellow balls.

### Step 1: Determine the total number of balls
The bag contains:
- 7 blue balls
- 5 yellow balls

**Total number of balls = 7 + 5 = 12**

### Step 2: Calculate the total number of ways to choose 2 balls from 12
We can use the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \) to find the total number of ways to choose 2 balls from 12.

\[ \text{Total ways to choose 2 balls} = C(12, 2) = \frac{12!}{2!(12-2)!} = \frac{12!}{2! \cdot 10!} \]

Calculating this gives:

\[ C(12, 2) = \frac{12 \times 11}{2 \times 1} = 66 \]

### Step 3: Calculate the number of ways to choose 2 blue balls
Since we want to find the probability that none of the selected balls are yellow, we will only consider the blue balls. We need to find the number of ways to choose 2 blue balls from the 7 blue balls.

\[ \text{Ways to choose 2 blue balls} = C(7, 2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2! \cdot 5!} \]

Calculating this gives:

\[ C(7, 2) = \frac{7 \times 6}{2 \times 1} = 21 \]

### Step 4: Calculate the probability that none of the selected balls are yellow
Now, we can find the probability that none of the selected balls are yellow by dividing the number of favorable outcomes (choosing 2 blue balls) by the total outcomes (choosing any 2 balls).

\[ \text{Probability (none are yellow)} = \frac{\text{Ways to choose 2 blue balls}}{\text{Total ways to choose 2 balls}} = \frac{C(7, 2)}{C(12, 2)} = \frac{21}{66} \]

### Step 5: Simplify the probability
We can simplify \( \frac{21}{66} \):

\[ \frac{21}{66} = \frac{7}{22} \]

### Final Answer
The probability that none of the selected balls are yellow is \( \frac{7}{22} \).

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