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 that;

 \(P(EA)=0.45\) ,\(P(EB)=0.55\) 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,

\(P(A' ∩ B') = P(A') × P(B')\)

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