Question

There are two cash counters A and B for placing orders in a college canteen. Let \(EA\) be the event that there is a queue at counter A and \(EB\) denotes the event that there is a queue at counter B. If \(P(EA)=0.45\) ,\(P(EB)=0.55\) and \(P(EA∩EB)=0.25\),then the probability that there is no queue at both the counters is:

• A. \(0.75\)
• B. \(0.15\)
• C. \(0.25\)
• D. \(0.2\)
• E. \(1.25\)

Answer 1

The Correct Option is C

Given:

• \( P(E_A) = 0.45 \) (probability of queue at counter A)
• \( P(E_B) = 0.55 \) (probability of queue at counter B)
• \( P(E_A \cap E_B) = 0.25 \) (probability of queue at both counters)

 

We need to find the probability that there is no queue at both counters, which is \( P(E_A^c \cap E_B^c) \).

Step 1: Use the probability of union: \[ P(E_A \cup E_B) = P(E_A) + P(E_B) - P(E_A \cap E_B) = 0.45 + 0.55 - 0.25 = 0.75 \]

Step 2: The probability of no queue at both counters is the complement of the union: \[ P(E_A^c \cap E_B^c) = 1 - P(E_A \cup E_B) = 1 - 0.75 = 0.25 \]

However, let's verify using De Morgan's Law: \[ P(E_A^c \cap E_B^c) = P((E_A \cup E_B)^c) = 1 - P(E_A \cup E_B) = 0.25 \]

The correct answer is (C) 0.25.

Answer 2

Given

 \(P(EA)=0.45\) ,\(P(EB)=0.55\) and \(P(EA∩EB)=0.25\)

for event \(EA\) and \(EB\) 

To find \(P(A' ∩ B')\), which is the probability that neither counter has a queue. (Where,  No queue at counter A ⇢ A' and

 No queue at counter B ⇢ B'.)

\(P(A ∪ A') = 1\) (The probability that either there is a queue at A or there is no queue at A is certain, so P(A ∪ A') = 1)

\(P(B ∪ B') = 1\) (The probability that either there is a queue at B or there is no queue at B is certain, so P(B ∪ B') = 1)

Now, by using relations of probability; 

\(P(A ∪ A') = P(A) + P(A') - P(A ∩ A')\)

\(P(B ∪ B') = P(B) + P(B') - P(B ∩ B')\)

putting numerical values in the above relation. given as per the question we get,

\(P(A ∪ A') = P(A) + P(A') - P(A ∩ A')\)

\(1 = 0.45 + P(A') - 0\)

(Because there is no intersection of \(A\) and \(A'\), since these are mutually exclusive events)

\(P(A') = 1 - 0.45\)

\(⇒P(A') = 0.55\)

Similarly, \(P(B') = 1 - 0.55\)

\(P(B') = 0.45\)

Now we need to find the \(P(A' ∩ B')\) as,

\(P(A' ∩ B') = P(A') × P(B')\)

\(⇒P(A' ∩ B') = 0.55 × 0.45 P(A' ∩ B') = 0.2475≈0.25\)