Question

If $P(A) = 0.3,P(B) = 0.4,P(C) = 0.8, P(AB) = 0.08,P(AC) = 0.28, P(ABC) = 0.09, P(A + B + C) \geqslant 0.75$ and $P(BC) = x$ then
A). $0.23 \leqslant x \leqslant 0.48$
B). $0.32 \leqslant x \leqslant 0.84$
C). $0.25 \leqslant x \leqslant 0.73$
D). None of these

Answer 1

Here we are asked to find the range of $P(B \cap C)$ . We know that the probability of an event will be less than or equal to one. With this, we will find the lower bound by using the probability formula. The formula that we choose has to have the terms that we have and the terms that we require.
Formula Used: Formula that we will be using in this problem: $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$

Complete step-by-step solution:
We aim to find the range of the term $x$ that is $P(B \cap C)$ . We are also given some values they are $P(A) = 0.3,P(B) = 0.4,P(C) = 0.8,P(AC) = 0.28,P(ABC) = 0.09,P(A + B + C) \geqslant 0.75$ and$P(BC) = x$ .
We know that the probability of an event will be between zero to one (less than or equal to one).
Form the given data we have $P(A \cup B \cup C) \geqslant 0.75$ and from the above statement, we can write$0.75 \leqslant P(A \cup B \cup C) \leqslant 1$ . But we need to find the range of the term $x$ that is $P(B \cap C)$ . We will change the above range by using the formula,
$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$ as
$0.75 \leqslant P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \leqslant 1$
We already have the values of$P(A),P(B),P(C),P(A \cap B),P(A \cap C),P(A \cap B \cap C)\& P(A \cup B \cup C)$. Now we will substitute those values in the formula.
$0.75 \leqslant 0.3 + 0.4 + 0.8 - 0.08 - 0.28 - P(B \cap C) + 0.09 \leqslant 1$
Let us simplify the above inequality.
$0.75 \leqslant 1.23 - P(B \cap C) \leqslant 1$
Now let us subtract $1.23$ throughout the inequality.
$0.75 - 1.23 \leqslant 1.23 - P(B \cap C) - 1.23 \leqslant 1 - 1.23$
On simplifying this we get
$- 0.48 \leqslant - P(B \cap C) \leqslant - 0.23$
Now let us multiply by $- 1$ throughout the above inequality.
$- 0.48 \times - 1 \leqslant - P(B \cap C) \times - 1 \leqslant - 0.23 \times - 1$
On simplifying this we get
$0.23 \leqslant P(B \cap C) \leqslant 0.48$
Here the position of the range changes since $0.48 \geqslant 0.23$
And from the given data we have that $P(B \cap C) = x$ on substituting this we get
$0.23 \leqslant x \leqslant 0.48$
Thus, we have found the range of the term $x$. It lies between $0.23\& 0.48$ .
Now let us see the options, option (a) $0.23 \leqslant x \leqslant 0.48$ is the correct option as we got the same range in our calculation above.
Option (b) $0.32 \leqslant x \leqslant 0.84$ is an incorrect option as we got the range$0.23 \leqslant x \leqslant 0.48$ in our calculation.
Option (c) $0.25 \leqslant x \leqslant 0.73$ is an incorrect option as we got the range $0.23 \leqslant x \leqslant 0.48$ in our calculation.
Option (d) None of these is an incorrect option as we got the option (a) as the correct answer.
Hence, option (a) $0.23 \leqslant x \leqslant 0.48$ is the correct answer.

Note: Here the terms P(AB) is denoted as $P(A \cap B)$, P(AC) is denoted as $P(A \cap C)$, P(BC) is denoted as $P(B \cap C)$, P(ABC) is denoted as $P(A \cap B \cap C)$, and P(A+B+C) is denoted as $P(A \cup B \cup C)$. Then we have to choose the appropriate formula so that it will help us to find the value that we require.