One ticket is selected at random from 50 tick | vedclass

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

Question

(B) Let $S$ be the sample space of $50$ tickets: $S = \{00, 01, 02, \ldots, 49\}$,so $n(S) = 50$.Let $E_1$ be the event that the sum of the digits is

Vedclass · Accepted Answer

$\frac{1}{14}$

Vedclass · Answer

(B) Let $S$ be the sample space of $50$ tickets: $S = \{00, 01, 02, \ldots, 49\}$,so $n(S) = 50$.Let $E_1$ be the event that the sum of the digits is $8$: $E_1 = \{08, 17, 26, 35, 44\}$,so $n(E_1) = 5$.Let $E_2$ be the event that the product of the digits is zero. This occurs if at least one digit is $0$: $E_2 = \{00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40\}$,so $n(E_2) = 14$.The intersection $E_1 \cap E_2$ is the set of tickets where the sum is $8$ and the product is $0$: $E_1 \cap E_2 = \{08\}$,so $n(E_1 \cap E_2) = 1$.The conditional probability is $P(E_1 | E_2) = \frac{n(E_1 \cap E_2)}{n(E_2)} = \frac{1}{14}$.

One ticket is selected at random from $50$ tickets numbered $\{00, 01, 02, \ldots, 49\}$. Then the probability that the sum of the digits on the selected ticket is $8$,given that the product of these digits is zero,is

Similar Questions

Solve the differential equation given below:
$\frac{x dy}{dx} = y + \sqrt{x^2 + y^2}$

The equation of the curve which passes through point $(1,0)$ and has a tangent with slope $1+\frac{y}{x}+\left(\frac{y}{x}\right)^{2}$ is

The solution of the differential equation $y \frac{dy}{dx} = x \left[ \frac{y^2}{x^2} + \frac{\phi(y^2/x^2)}{\phi'(y^2/x^2)} \right]$ is (where $c$ is a constant):

The substitution $y = z^{\alpha}$ transforms the differential equation $(x^2y^2 - 1)dy + 2xy^3dx = 0$ into a homogeneous differential equation for

The general solution of the differential equation $\frac{dy}{dx} = \frac{x+y}{x-y}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module