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If P(A) = 0.65, P(B)=0.80, then $P(A \cap B)$ lies in the interval.
(A). [0.30, 0.80]
(B). [0.35, 0.75]
(C). [0.4, 0.70]
(D). [0.45, 0.65]

seo-qna
Last updated date: 22nd Mar 2024
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MVSAT 2024
Answer
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Hint: When probabilities of events are given then we try to solve the question by formulas. Also, we have to tell the interval so we’ll try to analyse the minimum and maximum value of the probability. While solving the probability we should remember that $P(A \cup B) = 1$.

Complete step-by-step answer:
Here we have given probability if an event A is 0.65 and event B is 0.80. And we have asked the probability of intersection of both the sets. Since we need to tell the interval, we’ll see what would be the minimum value and what would be the maximum value of the intersection of both the events.
We know that, $P(A \cap B) \leqslant \min \{ P(A),P(B)\} $
On putting the values in above formula, we get,
$
  P(A \cap B) \leqslant \min \{ P(A),P(B)\} \\
   \Rightarrow P(A \cap B) \leqslant \min \{ 0.65,0.80\} \\
   \Rightarrow P(A \cap B) \leqslant 0.65 \\
$
From this we have got our upper bound that is the maximum value of probability of intersection of events.
For the lower bound or minimum value, we’ll use,
$P(A \cap B) = P(A) + P(B) - P(A \cup B)$
Also, we need to find the minimum value of $P(A \cap B)$. For $P(A \cap B)$ to be minimum $P(A \cup B)$ has to be maximum and we know the maximum value of any probability is 1. Now, on putting all the values we get,
$
  P(A \cap B) = P(A) + P(B) - P(A \cup B) \\
   \Rightarrow P(A \cap B) \geqslant 0.65 + 0.80 - 1 \\
   \Rightarrow P(A \cap B) \geqslant 1.45 + - 1 \\
   \Rightarrow P(A \cap B) \geqslant 0.45 \\
$
From this we have got our minimum value of $P(A \cap B)$.
So, $0.45 \leqslant P(A \cap B) \leqslant 0.65$ which can also be expressed as [0.45, 0.65]
Hence, Option (D) is the correct option.

Note: Probability is the core phenomena of predictive analysis. It has several variations. For example, in this question Instead of 2 sets they can also give n number of sets. In such questions we need to generalise the basic formula of probability for n events. For example, $P\left( {\bigcup\limits_{i = 1}^n {{X_i}} } \right) = 1$.