The probability that a candidate selected in competitive examinations of B.S.F., C.D.S., Bank P.O. and a, b and c respectively. Of these examinations, a candidate has 70% chance of selection in at least one, 50% chance of selection in at least two and 30% chance of selection in exactly two examinations. If `a+b+c=I/m`, then find `l+m` if LCM `(m)=1`.

Options

A. `1//5`
B. 12
C. `1//2`
D. 5

Answer

To solve the problem step by step, we will analyze the given probabilities and use them to find the values of \(a\), \(b\), and \(c\).

### Step 1: Understand the Given Probabilities
We are given:
- The probability of selection in at least one examination is 70% (or 0.7).
- The probability of selection in at least two examinations is 50% (or 0.5).
- The probability of selection in exactly two examinations is 30% (or 0.3).

### Step 2: Calculate the Probability of Exactly One Examination
From the probabilities given, we can find the probability of exactly one examination being passed.

Using the formula:
\[ P(\text{at least 1}) = P(\text{exactly 1}) + P(\text{exactly 2}) + P(\text{exactly 3}) \]
We know that:
\[ P(\text{at least 1}) = 0.7 \]
\[ P(\text{at least 2}) = P(\text{exactly 2}) + P(\text{exactly 3}) = 0.5 \]
Thus, we can find \(P(\text{exactly 1})\):
\[ P(\text{exactly 1}) = P(\text{at least 1}) - P(\text{at least 2}) = 0.7 - 0.5 = 0.2 \]

### Step 3: Calculate the Probability of Exactly Three Examinations
Now, we can find the probability of exactly three examinations being passed. Since we know that:
\[ P(\text{exactly 3}) = P(\text{at least 2}) - P(\text{exactly 2}) = 0.5 - 0.3 = 0.2 \]

### Step 4: Set Up the Venn Diagram Equation
Let \(a\), \(b\), and \(c\) be the probabilities of passing the BSF, CDS, and Bank PO examinations respectively. The total probability can be expressed as:
\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \]
Where:
- \(P(A) = a\)
- \(P(B) = b\)
- \(P(C) = c\)
- \(P(A \cap B) + P(A \cap C) + P(B \cap C) = 2 \times P(\text{exactly 2}) = 2 \times 0.3 = 0.6\)
- \(P(A \cap B \cap C) = P(\text{exactly 3}) = 0.2\)

### Step 5: Substitute the Values
Now substituting the values into the equation:
\[ P(A \cup B \cup C) = a + b + c - 0.6 + 0.2 \]
Since \(P(A \cup B \cup C) = 0.7\), we have:
\[ 0.7 = a + b + c - 0.4 \]
Thus:
\[ a + b + c = 0.7 + 0.4 = 1.1 \]

### Step 6: Relate to Given Equation
We are given that:
\[ a + b + c = \frac{L}{M} \]
From our calculation, we have:
\[ \frac{L}{M} = 1.1 \]
This can be rewritten as:
\[ L = 1.1M \]
To express \(L\) and \(M\) in integers, we can multiply both sides by 10:
\[ 10L = 11M \]
Thus, we can take \(L = 11\) and \(M = 10\).

### Step 7: Find \(L + M\)
Now, we need to find \(L + M\):
\[ L + M = 11 + 10 = 21 \]

### Final Answer
Thus, the required answer is:
\[ \boxed{21} \]