There are five mangoes and six organges in a bucket what will be probability of Picking up four fruits which contains at least two orange ?

Options

A. A)`(53)/(66)`
B. B)`(43)/(66)`
C. C)`(59)/(66)`
D. D)`(49)/(66)`

Answer

To solve the problem of finding the probability of picking up 4 fruits from a bucket containing 5 mangoes and 6 oranges, with the condition that at least 2 of the fruits must be oranges, we can break down the solution into clear steps.

### Step 1: Understand the Total Number of Fruits
We have a total of:
- 5 mangoes
- 6 oranges

Total fruits = 5 + 6 = 11 fruits.

### Step 2: Determine the Favorable Outcomes
We need to find the number of ways to pick 4 fruits that contain at least 2 oranges. We can break this down into three cases:
1. Case 1: 2 oranges and 2 mangoes
2. Case 2: 3 oranges and 1 mango
3. Case 3: 4 oranges

### Step 3: Calculate Each Case
**Case 1: 2 oranges and 2 mangoes**
- The number of ways to choose 2 oranges from 6: \( \binom{6}{2} \)
- The number of ways to choose 2 mangoes from 5: \( \binom{5}{2} \)

Total ways for Case 1:
\[ \text{Ways} = \binom{6}{2} \times \binom{5}{2} \]

**Case 2: 3 oranges and 1 mango**
- The number of ways to choose 3 oranges from 6: \( \binom{6}{3} \)
- The number of ways to choose 1 mango from 5: \( \binom{5}{1} \)

Total ways for Case 2:
\[ \text{Ways} = \binom{6}{3} \times \binom{5}{1} \]

**Case 3: 4 oranges**
- The number of ways to choose 4 oranges from 6: \( \binom{6}{4} \)

Total ways for Case 3:
\[ \text{Ways} = \binom{6}{4} \]

### Step 4: Sum the Favorable Outcomes
Now, we will sum the number of ways from all three cases:
\[ \text{Total Favorable Outcomes} = \binom{6}{2} \times \binom{5}{2} + \binom{6}{3} \times \binom{5}{1} + \binom{6}{4} \]

### Step 5: Calculate Total Outcomes
The total number of ways to choose any 4 fruits from 11 is given by:
\[ \text{Total Outcomes} = \binom{11}{4} \]

### Step 6: Calculate the Probability
The probability \( P \) of picking 4 fruits with at least 2 oranges is given by:
\[ P = \frac{\text{Total Favorable Outcomes}}{\text{Total Outcomes}} \]

### Step 7: Substitute Values and Simplify
Now we will substitute the values of the combinations:
- \( \binom{6}{2} = 15 \)
- \( \binom{5}{2} = 10 \)
- \( \binom{6}{3} = 20 \)
- \( \binom{5}{1} = 5 \)
- \( \binom{6}{4} = 15 \)
- \( \binom{11}{4} = 330 \)

Calculating:
1. Case 1: \( 15 \times 10 = 150 \)
2. Case 2: \( 20 \times 5 = 100 \)
3. Case 3: \( 15 \)

Total Favorable Outcomes:
\[ 150 + 100 + 15 = 265 \]

Total Outcomes:
\[ 330 \]

### Final Probability Calculation
\[ P = \frac{265}{330} = \frac{53}{66} \]

Thus, the required probability is:
\[ \boxed{\frac{53}{66}} \]