A MNC has two Grids - Grid I and Grid II. Out of 5 Directors and 4 General Managers of Grid I, one person is transferred to Grid II having already 3 Directors and 7 General Managers. If, one person superannuates from Grid II, then the probability that this person was a director is

Options

A. `32/99`
B. `4/45`
C. `20/99`
D. `3/10`

Answer

To solve the problem, we need to find the probability that a person who superannuates from Grid II is a director. We'll break this down step by step.

### Step 1: Determine the initial counts in both grids
- **Grid I**:
- Directors = 5
- General Managers = 4
- Total in Grid I = 5 + 4 = 9

- **Grid II**:
- Directors = 3
- General Managers = 7
- Total in Grid II = 3 + 7 = 10

### Step 2: Analyze the transfer from Grid I to Grid II
One person from Grid I is transferred to Grid II. This person can either be a director or a general manager. We will consider both cases.

### Step 3: Case 1 - A Director is Transferred
- If a director is transferred:
- New counts in Grid I:
- Directors = 5 - 1 = 4
- General Managers = 4
- Total in Grid I = 4 + 4 = 8
- New counts in Grid II:
- Directors = 3 + 1 = 4
- General Managers = 7
- Total in Grid II = 4 + 7 = 11

### Step 4: Case 2 - A General Manager is Transferred
- If a general manager is transferred:
- New counts in Grid I:
- Directors = 5
- General Managers = 4 - 1 = 3
- Total in Grid I = 5 + 3 = 8
- New counts in Grid II:
- Directors = 3
- General Managers = 7 + 1 = 8
- Total in Grid II = 3 + 8 = 11

### Step 5: Calculate the probability of selecting a director from Grid II
Now we need to find the probability of selecting a director from Grid II after the transfer.

#### Case 1 Probability:
- Total in Grid II = 11 (4 directors + 7 general managers)
- Probability of selecting a director = Number of Directors / Total = 4 / 11

#### Case 2 Probability:
- Total in Grid II = 11 (3 directors + 8 general managers)
- Probability of selecting a director = Number of Directors / Total = 3 / 11

### Step 6: Calculate the overall probability
Now we need to find the probability of each case happening:
- Probability of transferring a director from Grid I = 5/9
- Probability of transferring a general manager from Grid I = 4/9

Now, we can find the total probability of selecting a director from Grid II using the law of total probability:

\[ P(\text{Director}) = P(\text{Director transferred}) \times P(\text{Director from Grid II | Director transferred}) + P(\text{GM transferred}) \times P(\text{Director from Grid II | GM transferred}) \]

Substituting the values:

\[ P(\text{Director}) = \left(\frac{5}{9} \times \frac{4}{11}\right) + \left(\frac{4}{9} \times \frac{3}{11}\right) \]

Calculating each term:

1. \( \frac{5}{9} \times \frac{4}{11} = \frac{20}{99} \)
2. \( \frac{4}{9} \times \frac{3}{11} = \frac{12}{99} \)

Adding these together:

\[ P(\text{Director}) = \frac{20}{99} + \frac{12}{99} = \frac{32}{99} \]

### Final Answer:
The probability that the person who superannuates from Grid II is a director is \( \frac{32}{99} \).

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