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

Answer: (C)

$0.25$

Answer 2

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.)