A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective . If we choose two fruits at random, what is the probability that either both are oranges or both are non defective?

Options

A. A. `136/345`
B. B. `17/87`
C. C. `316/435`
D. D. `158/435`

Answer

To solve the problem, we need to find the probability that either both fruits chosen are oranges or both fruits are non-defective.

### Step 1: Determine the total number of fruits.
The total number of fruits in the basket is:
- Apples: 10
- Oranges: 20

Total fruits = 10 + 20 = 30

**Hint:** Count all the fruits in the basket to find the total.

### Step 2: Calculate the total ways to choose 2 fruits from 30.
The total number of ways to choose 2 fruits from 30 is given by the combination formula:
\[ \text{Total ways} = \binom{30}{2} = \frac{30 \times 29}{2} = 435 \]

**Hint:** Use the combination formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \) to find the number of ways to choose items.

### Step 3: Calculate the number of ways to choose 2 oranges.
The number of oranges in the basket is 20. The number of ways to choose 2 oranges is:
\[ \text{Ways to choose 2 oranges} = \binom{20}{2} = \frac{20 \times 19}{2} = 190 \]

**Hint:** Focus on the specific fruit type you are interested in (oranges in this case).

### Step 4: Calculate the number of non-defective fruits.
The total number of defective fruits:
- Defective apples: 3
- Defective oranges: 5

Total defective fruits = 3 + 5 = 8

Thus, the number of non-defective fruits is:
\[ \text{Non-defective fruits} = 30 - 8 = 22 \]

**Hint:** Subtract the defective fruits from the total to find the non-defective ones.

### Step 5: Calculate the number of ways to choose 2 non-defective fruits.
The number of ways to choose 2 non-defective fruits is:
\[ \text{Ways to choose 2 non-defective fruits} = \binom{22}{2} = \frac{22 \times 21}{2} = 231 \]

**Hint:** Again, use the combination formula for non-defective fruits.

### Step 6: Calculate the number of ways to choose 2 defective oranges.
The number of defective oranges is 5. The number of ways to choose 2 defective oranges is:
\[ \text{Ways to choose 2 defective oranges} = \binom{5}{2} = \frac{5 \times 4}{2} = 10 \]

**Hint:** This step focuses on the defective fruits specifically.

### Step 7: Use the principle of inclusion-exclusion to find the probability.
We need to find the probability that either both fruits are oranges or both are non-defective. This can be expressed as:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Where:
- \( P(A) \) is the probability of choosing 2 oranges.
- \( P(B) \) is the probability of choosing 2 non-defective fruits.
- \( P(A \cap B) \) is the probability of choosing 2 non-defective oranges.

Calculating each:
\[ P(A) = \frac{190}{435} \]
\[ P(B) = \frac{231}{435} \]
\[ P(A \cap B) = \frac{10}{435} \]

Now substituting back into the inclusion-exclusion formula:
\[ P(A \cup B) = \frac{190}{435} + \frac{231}{435} - \frac{10}{435} = \frac{411}{435} \]

### Step 8: Simplify the probability.
The final probability can be simplified:
\[ P(A \cup B) = \frac{411}{435} \approx 0.9448 \]

**Hint:** Always check if the fraction can be simplified further.

### Final Answer:
The probability that either both fruits are oranges or both are non-defective is \( \frac{411}{435} \).