Question

The probability that at least one of $A$ and $B$ occurs is $0.6$ and probability that they occur simultaneously is $0.3$, then $P(A') + P(B')$ is:
A)$0.9$
B) $1.15$
C) $1.1$
D) $1.2$

Answer 1

Hint: We can solve the above question with the help of the concept ‘rule of addition’ where union and intersection of $A$ and $B$ events, used in probability as shown in the formula we are going to use in this question.

Complete step-by-step answer:
Given:
The probability that at least one of $A$ and $B$ occurs,$P(A \cup B) = 0.6$
The Probability that $A$ and $B$ occur simultaneously,$P(A \cap B) = 0.3$
Probability: The chance of an event occurring is known as probability. The probability of an event will be in between the range of $0$ and $1$, in probability $0$ shows that there is no possibility of occurring an event and $1$ shows that possibility of occurring an event is $100\% $
To find the $P(A') + P(B')$ we will use below probability formula
\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]
We will put the given values in the formula
$0.6 = P(A) + P(B) - 0.3......(1)$
We know that $P(A) = 1 - P(A')$ and $P(B) = 1 - P(B')$ which we will put in equation $(1)$
\[0.9 = 1 - P(A') + 1 - P(B')\]
We will take numerical terms on left side reaming on right side
\[0.9 - 2 = - (P(A') + P(B'))\]
$P(A') + P(B') = 1.1$
Hence, we get the value of $P(A') + P(B')$ equals to $1.1$, which means the C) option is the right option.

Additional information:
Addition rule of probability: If events A and B come from the same sample area, the probability that event A and event B occur or event A or event B occurs is equal to the probability that event A occurs plus the probability that event B occurs minus the probability that both events A and B occur simultaneously. Which we can write in equation as: \[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]


Note: We should keep remembering that the total probability of happening and not happening an event is equal to one. So, we can get the probability of not happening by subtracting happening events from one.