Question

Probability is $0.45$ that a dealer will sell at least $20$ television sets during a day, and the probability is $0.74$ that he will sell less that $24$ televisions. The probability that he will sell $20,21,22$ or $23$ televisions during the day, is

• A. $0.19$
• B. $0.32$
• C. $0.21$
• D. None of these

Answer 1

Answer: (A)

$0.19$

Let $A$ be the event that the sale is at least $20$ televisions, i.e. $20,21,22,...$
and $B$ be the event that sale is less than $24$ i.e. $\mathrm{0.1.2.3...23}$.
Then $A\cap B$ will denote the sale of $20,21,22$ and $23$ televisions,
We are given $P\left(A\right)=0.45$ and $P\left(B\right)=0.74$.
It is required to find $P(A\cap B)$.
Also $P(A\cup B)=P($ sale of $0,1,2,3,...,20,21,22,23$ televisions$)=P(S)=1$
From addition rule, required probability is
$P(A\cap B)=P\left(A\right)+P\left(B\right)-P(A\cup B)=0.45+0.74-1=0.19$