A random variable X has the following probability distribution: X… - Vidyadip

Question

A random variable X has the following probability distribution:

X:	1	2	3	4	5	6	7	8
P(X):	0.15	0.23	0.12	0.10	0.20	0.08	0.07	0.05

Find the events E = {X : X is a prime number}, F{X : X < 4}, the probability $\text{P}(\text{E}\cup\text{F})$ is:

✓

Answer

0.77

Solution:
P(E) = P(2) + P(3) + P(5) + P(7)
P(E) = 0.23 + 0.12 + 0.20 + 0.07
P(E) = 0.62
And
P(F) = P(1) + P(2) + P(3)
P(F) = 0.15 + 0.23 + 0.12
P(F) = 0.5
Also,
$\text{P}(\text{E}\cap\text{F})=\text{P}(2)+\text{P}(3)$
$\text{P}(\text{E}\cap\text{F})=0.23+0.12$
$\text{P}(\text{E}\cap\text{F})=0.35$
$\text{P}(\text{E}\cup\text{F})=\text{P}(\text{E})+\text{P(F)}-\text{P}(\text{E}\cap\text{F})$
$\text{P}(\text{E}\cup\text{F})=0.62+0.5-0.35$
$\text{P}(\text{E}\cup\text{F})=0.77$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Probability→Probability — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Area lying first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the line $x = 0$ and $x = 2,$ is:

If $\vec{\text{a}}.\vec{\text{b}}=\vec{\text{a}}.\vec{\text{c}}$ and $\vec{\text{a}}\times\vec{\text{b}}=\vec{\text{a}}\times\vec{\text{c}}.\vec{\text{a}}\neq0,$ then:

$\vec{\text{b}}=\vec{\text{c}}$
$\vec{\text{b}}=\vec{0}$
$\vec{\text{b}}+\vec{\text{c}}=\vec{0}$
$\text{None of these}$

A five-digit number is written down at raddom. The probability that the number is divisible by 5, and no two consecutive digits are identical, is:

If A = [1, 2, 3], then the set of elements of A is:

[1, 2, 3]
[2, 0]
Only 2
None of these

If $a * b = a^2 + b^2,$ then the value of $(4 * 5) * 3$ is:

If $\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}\Big\}=\alpha,$ then $x^2 =$

The corner points of the feasible region determined by the system of linear inequalities are (0, 0), (4, 0), (2, 4) and (0, 5). If the maximum value of z = ax + by, where a, b > 0 occurs at both (2, 4) and (4, 0), then:

a = 2b
2a = b
a = b
3a = b

If $\theta$ is the angle between the vectors $2\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$ and $3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}},$ then $\sin\theta=$

$\frac{2}{3}$
$\frac{2}{\sqrt{7}}$
$\frac{\sqrt{2}}{7}$
$\sqrt{\frac{2}{7}}$

The corner points of the feasible region determined by the following system of linear inequalities:
2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5).
Let Z = px + qy, where p.q > 0.
Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is:

Area lying first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the line $x = 0$ and $x = 2,$ is: