Finding out the probability distribution of numbers from -3 to 3

Question

I am really able to solve problems with the help of all the volunteers here.

a big thanks to every one..

Please explain this problem..

A random variable 'X' takes the values -3,-2,-1,0,1,2,3. Such that P(X=0) = P(X<0) = P(X>0) and P(X=-3) = P(X=-2) = P(X=-1) and P(X=3) = P(X=2) = P(X=1). Obtain the distribution of X.

I have found that P(X=0) = 1/7; and now how to proceed...

Let the probability that $x=-3$ be $a$. Then everybody has probability $a$, except $0$, who has probability $3a$. — André Nicolas, Jul 28 '14 at 15:33

score 1 · Accepted Answer · answered Jul 28 '14 at 15:32

1

You've found it incorrectly. The first condition tells us that $$ P(X = 0) = P(X > 0) = P(X < 0) =\frac13 $$ since the sum of these three probabilities is $1$. For other $6$ points, probabilities are $\frac19$.

answered Jul 28 '14 at 15:32

SBF

36,041

Okay, so P(X=0)+P(X<0)+P(X<0) = 1
Thank you..
– Ganesh Vellanki Jul 28 '14 at 15:48
@VellankiGaneshBabu: glad to be of help – SBF Jul 28 '14 at 15:50

score 0 · Answer 2 · answered Jul 28 '14 at 15:37

0

Hint

Set $p_{-3}=p_{-2}=p_{-1}=a$ and $p_{3}=p_{2}=p_{1}=b$ then:

$$3a+p_{0}+3b=1$$

$$3a=p_{0}=3b$$

answered Jul 28 '14 at 15:37

drhab

151,093

Thank you @drhab, a different approach and how to find out the value of P(X=0)?? – Ganesh Vellanki Jul 28 '14 at 16:11
Here $P(X=0)$ is denoted as $p_0$ and with these equations you can find $3p_0=3a+p_0+3b=1$ hence $p_0=\frac{1}{3}$. – drhab Jul 29 '14 at 06:45

Finding out the probability distribution of numbers from -3 to 3

2 Answers2