Let A and B be two independent events. The pr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that $A$ and $B$ are independent events, $P(A \cap B) = P(A)P(B) = \frac{1}{12} \dots (1)$ The probability that neither $A$ nor $B$ happens

Vedclass · Accepted Answer

$P(A)=\frac{1}{3}, P(B)=\frac{1}{4}$

Vedclass · Answer

(A) Given that $A$ and $B$ are independent events, $P(A \cap B) = P(A)P(B) = \frac{1}{12} \dots (1)$ The probability that neither $A$ nor $B$ happens is $P(A' \cap B') = \frac{1}{2}$. By De Morgan's Law, $P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) = \frac{1}{2}$, so $P(A \cup B) = \frac{1}{2}$. Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, we get: $P(A) + P(B) - \frac{1}{12} = \frac{1}{2} \implies P(A) + P(B) = \frac{1}{2} + \frac{1}{12} = \frac{7}{12} \dots (2)$ From $(1)$ and $(2)$, $P(A)$ and $P(B)$ are roots of the quadratic equation $x^2 - (P(A)+P(B))x + P(A)P(B) = 0$. $x^2 - \frac{7}{12}x + \frac{1}{12} = 0 \implies 12x^2 - 7x + 1 = 0$. Factoring the quadratic: $12x^2 - 4x - 3x + 1 = 0 \implies 4x(3x - 1) - 1(3x - 1) = 0 \implies (4x - 1)(3x - 1) = 0$. Thus, $x = \frac{1}{4}$ or $x = \frac{1}{3}$. Therefore, the probabilities are $P(A) = \frac{1}{3}$ and $P(B) = \frac{1}{4}$ (or vice versa).

Box	White	Black
$B_1$	$1$	$2$
$B_2$	$3$	$1$
$B_3$	$2$	$3$

Let $A$ and $B$ be two independent events. The probability that both $A$ and $B$ happen is $\frac{1}{12}$ and the probability that neither $A$ nor $B$ happens is $\frac{1}{2}$. Then

Similar Questions

$A$ number $n$ is chosen at random from the natural numbers $2$ to $1001$. The probability that $n$ is a number that leaves remainder $1$ when divided by $7$ is:

One card is drawn from each of two ordinary packs of $52$ cards. The probability that at least one of them is an ace of hearts is:

The events $A$ and $B$ have probabilities $0.25$ and $0.50$, respectively. The probability that both $A$ and $B$ occur simultaneously is $0.14$. Then, the probability that neither $A$ nor $B$ occurs is:

$A$ card is drawn at random from a pack of $100$ cards numbered $1$ to $100$. The probability of drawing a number which is a perfect square is

Boxes $B_1, B_2$,and $B_3$ contain balls as given below:
Box White Black
$B_1$ $1$ $2$
$B_2$ $3$ $1$
$B_3$ $2$ $3$

One ball is drawn at random from each box. Then,among the balls drawn,the probability that two are black and one is white,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module