Let X denote the number of hours you study on a Sunday. Also it is known that P(X=x) = begincases 0.1 ,& textif x = 0 kx ,& textif x=1 textor 2 k(5-x) ,& textif x = 3 textor 4 0 , textotherwise endcases where k is a constant. What is the probability that you study atleast two hours ?

Question

Let X denote the number of hours you study on a Sunday. Also it is known that P(X=x) = begincases 0.1 ,& textif x = 0 kx ,& textif x=1 textor 2 k(5-x) ,& textif x = 3 textor 4 0 , textotherwise endcases where k is a constant. What is the probability that you study atleast two hours ? (A) 0.55 (B) 0.15 (C) 0.75 (D) 0.3

Tardigrade Admin · Accepted Answer

From the given information, we find that the probability distribution of

X

is

We know that

\sum p_{i} = 1

\Rightarrow 0.1 + k + 2 k + 2 k + k = 1

\Rightarrow 6 k = 1 - 0.1 = 0.9

\Rightarrow k = 0.15

Now,

P

(you study atleast two hours)

= P (X \geq 2)

= P (2) + P (3) + P (4) = 2 k + 2 k + k

= 5 k = 5 \times 0.15 = 0.75

Q. Let $X$ denote the number of hours you study on a Sunday. Also it is known that
$P (X = x) = ⎩ ⎨ ⎧ 0.1, k x, k (5 - x), 0, if x = 0 if x = 1 or 2 if x = 3 or 4 otherwise$
where $k$ is a constant.
What is the probability that you study atleast two hours ?

$0.55$

$0.15$

$0.75$

$0.3$

Q. Let X denote the number of hours you study on a Sunday. Also it is known that P(X=x)=⎩⎨⎧​0.1,kx,k(5−x),0,​if x=0if x=1or 2if x=3or 4otherwise ​ where k is a constant. What is the probability that you study atleast two hours ?

0.55

0.15

0.75

0.3

From the given information, we find that the probability distribution of X is We know that ∑pi​=1 ⇒0.1+k+2k+2k+k=1 ⇒6k=1−0.1=0.9 ⇒k=0.15 Now, P(you study atleast two hours) =P(X≥2) =P(2)+P(3)+P(4)=2k+2k+k =5k=5×0.15=0.75

Q. Let $X$ denote the number of hours you study on a Sunday. Also it is known that
$P (X = x) = ⎩ ⎨ ⎧ 0.1, k x, k (5 - x), 0, if x = 0 if x = 1 or 2 if x = 3 or 4 otherwise$
where $k$ is a constant.
What is the probability that you study atleast two hours ?

$0.55$

$0.15$

$0.75$

$0.3$