Math Problem Statement

. A discrete random variable Y has a pmf given by ( )   for 0,1,2,..... 4 3 P Y  y  c y  x Find the value of the constant c , PX  3 and PX  3

Solution

Problem Breakdown:

The problem states that YY is a discrete random variable with the probability mass function (pmf):

P(Y=y)=cy,for y=0,1,2,P(Y = y) = c y, \quad \text{for } y = 0, 1, 2, \dots

Additionally, the probabilities must satisfy the property of a probability mass function:

y=0P(Y=y)=1.\sum_{y=0}^\infty P(Y = y) = 1.

We will:

  1. Solve for the constant cc using the above condition.
  2. Compute P(X<3)P(X < 3) and P(X3)P(X \geq 3).

Step 1: Solve for cc

From the pmf, we write:

y=0P(Y=y)=y=0cy=1.\sum_{y=0}^\infty P(Y = y) = \sum_{y=0}^\infty c y = 1.

However, this series is infinite. It is common to cap the range of yy to some maximum value (e.g., 4 here, y=0,1,2,3,4y = 0, 1, 2, 3, 4). Thus, the summation becomes:

y=04cy=1.\sum_{y=0}^4 c y = 1.

Calculating the sum:

y=04y=0+1+2+3+4=10.\sum_{y=0}^4 y = 0 + 1 + 2 + 3 + 4 = 10.

So,

c10=1    c=110.c \cdot 10 = 1 \implies c = \frac{1}{10}.


Step 2: Compute P(X<3)P(X < 3)

For X<3X < 3, the possible values of YY are 0,1,0, 1, and 22. Thus:

P(X<3)=P(Y=0)+P(Y=1)+P(Y=2).P(X < 3) = P(Y = 0) + P(Y = 1) + P(Y = 2).

Using c=110c = \frac{1}{10}, we calculate:

P(Y=0)=1100=0,P(Y=1)=1101=110,P(Y = 0) = \frac{1}{10} \cdot 0 = 0, \quad P(Y = 1) = \frac{1}{10} \cdot 1 = \frac{1}{10}, P(Y=2)=1102=210.P(Y = 2) = \frac{1}{10} \cdot 2 = \frac{2}{10}.

So:

P(X<3)=0+110+210=310.P(X < 3) = 0 + \frac{1}{10} + \frac{2}{10} = \frac{3}{10}.


Step 3: Compute P(X3)P(X \geq 3)

For X3X \geq 3, the possible values of YY are 33 and 44. Thus:

P(X3)=P(Y=3)+P(Y=4).P(X \geq 3) = P(Y = 3) + P(Y = 4).

Using c=110c = \frac{1}{10}, we calculate:

P(Y=3)=1103=310,P(Y=4)=1104=410.P(Y = 3) = \frac{1}{10} \cdot 3 = \frac{3}{10}, \quad P(Y = 4) = \frac{1}{10} \cdot 4 = \frac{4}{10}.

So:

P(X3)=310+410=710.P(X \geq 3) = \frac{3}{10} + \frac{4}{10} = \frac{7}{10}.


Final Answers:

  1. c=110c = \frac{1}{10},
  2. P(X<3)=310P(X < 3) = \frac{3}{10},
  3. P(X3)=710P(X \geq 3) = \frac{7}{10}.

Questions:

  1. How does a probability mass function (pmf) differ from a probability density function (pdf)?
  2. Can this approach extend to cases where YY has an infinite range?
  3. What properties of cc ensure that the total probability is always 1?
  4. How would you compute E[Y]E[Y] (expected value) for this pmf?
  5. If P(Y=y)=cy2P(Y = y) = cy^2, how would the solution for cc change?

Tip:

Always verify that the sum of probabilities in a pmf equals 1, as this ensures it is a valid probability distribution.

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Math Problem Analysis

Mathematical Concepts

Probability
Discrete Random Variables
Probability Mass Function (PMF)

Formulas

Sum of probabilities in a PMF: ΣP(Y = y) = 1
Probability computations: P(A) = ΣP(Y = y) for y in A

Theorems

Normalization property of PMF: The sum of all probabilities must equal 1.

Suitable Grade Level

Grades 11-12