Q.The probability that at least one of the two events $A$ and $B$ occurs is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.3$, evaluate $P(A') + P(B')$.

Concept: Probability Complement Rule — $P(A') = 1 - P(A)$.

We are given:

• $P(A \cup B) = 0.6$ (at least one occurs)
• $P(A \cap B) = 0.3$

Step 1: Use the addition rule:

So,

Step 2: Now,

Step 3: Substitute the sum:

The problem asks for $P(A') + P(B')$, the sum of the probabilities of the complements of two events. A direct approach would require knowing $P(A)$ and $P(B)$ individually, but we are not given those. Instead, we are given two pieces of information:

• $P(A \cup B) = 0.6$ — the probability that at least one occurs.
• $P(A \cap B) = 0.3$ — the probability that both occur simultaneously.

The complement rule tells us that $P(A') = 1 - P(A)$ and $P(B') = 1 - P(B)$. So:

The problem reduces to finding $P(A) + P(B)$ from the given union and intersection. This is where the addition rule of probability comes in.

Rearranging:

Now substitute the given values:

1. $P(A \cup B) = 0.6$
2. $P(A \cap B) = 0.3$

So:

Therefore:

Method: Relating Complement Sums to the Addition Rule

Use this when you must find a combination like $P(A')+P(B')$ but are given only the union and intersection.

Steps

Step 1: Convert the complements first.

By the complement rule $P(A')=1-P(A)$ and $P(B')=1-P(B)$, so

The problem reduces to finding the sum $P(A)+P(B)$ — the individual values are not needed.

Step 2: Recover the sum from the addition rule.

Step 3: Substitute. Put the sum from Step 2 into the expression from Step 1. Recognising that only the combined quantity is required is what makes this quick.

Common Mistakes

Mistake 1: Writing $P(A')+P(B')=1-P(A\cup B)$.

Why it's wrong: complements do not combine that way; $P(A')+P(B')=2-[P(A)+P(B)]$. Correct approach: convert each complement separately, then find the sum $P(A)+P(B)$.

Mistake 2: Trying to find $P(A)$ and $P(B)$ individually.

Why it's wrong: the data fix only their sum, not each value. Correct approach: use $P(A)+P(B)=P(A\cup B)+P(A\cap B)=0.9$, which is all that is needed to get $1.1$.

Mistake 3: Dropping the overlap when recovering the sum.

Why it's wrong: $P(A)+P(B)=P(A\cup B)+P(A\cap B)$, so the intersection is added back, not ignored.