The probability distribution of a random variable X is given by

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The probability distribution of a random variable X is given by f(0) = 0.89, f(1) = 0.07, f(2) = 0.03, f(3) = 0.01. Then, P(0 < X < 2) + P(X = 3) =

The Correct Answer and Explanation is:

We are given the probability distribution of a discrete random variable XXX as:

  • f(0)=P(X=0)=0.89f(0) = P(X = 0) = 0.89f(0)=P(X=0)=0.89
  • f(1)=P(X=1)=0.07f(1) = P(X = 1) = 0.07f(1)=P(X=1)=0.07
  • f(2)=P(X=2)=0.03f(2) = P(X = 2) = 0.03f(2)=P(X=2)=0.03
  • f(3)=P(X=3)=0.01f(3) = P(X = 3) = 0.01f(3)=P(X=3)=0.01

We are asked to find:P(0<X<2)+P(X=3)P(0 < X < 2) + P(X = 3)P(0<X<2)+P(X=3)

Step-by-step Solution:

Step 1: Understand the expression P(0<X<2)P(0 < X < 2)P(0<X<2)

Since XXX is a discrete random variable, P(0<X<2)P(0 < X < 2)P(0<X<2) means the probability that XXX is greater than 0 but less than 2. The only integer value that satisfies this condition is X=1X = 1X=1. Therefore,P(0<X<2)=P(X=1)=0.07P(0 < X < 2) = P(X = 1) = 0.07P(0<X<2)=P(X=1)=0.07

Step 2: Add the probability P(X=3)P(X = 3)P(X=3)

From the given distribution:P(X=3)=0.01P(X = 3) = 0.01P(X=3)=0.01

Step 3: Add the two probabilities

P(0<X<2)+P(X=3)=0.07+0.01=0.08P(0 < X < 2) + P(X = 3) = 0.07 + 0.01 = 0.08P(0<X<2)+P(X=3)=0.07+0.01=0.08

Final Answer:

0.08\boxed{0.08}0.08​

Textbook-style Explanation

In probability theory, a discrete probability distribution lists the probabilities associated with each possible value that a discrete random variable can take. In this case, the random variable XXX can take on the values 0, 1, 2, and 3, with the respective probabilities:

  • P(X=0)=0.89P(X = 0) = 0.89P(X=0)=0.89
  • P(X=1)=0.07P(X = 1) = 0.07P(X=1)=0.07
  • P(X=2)=0.03P(X = 2) = 0.03P(X=2)=0.03
  • P(X=3)=0.01P(X = 3) = 0.01P(X=3)=0.01

We are interested in computing the sum of two probabilities: P(0<X<2)P(0 < X < 2)P(0<X<2) and P(X=3)P(X = 3)P(X=3). The expression 0<X<20 < X < 20<X<2 refers to the values of XXX that are greater than 0 and less than 2. Since XXX is a discrete variable, we only consider integer values. The only integer between 0 and 2 is 1. Therefore:P(0<X<2)=P(X=1)P(0 < X < 2) = P(X = 1)P(0<X<2)=P(X=1)

From the given distribution, we find:P(X=1)=0.07andP(X=3)=0.01P(X = 1) = 0.07 \quad \text{and} \quad P(X = 3) = 0.01P(X=1)=0.07andP(X=3)=0.01

Adding these probabilities gives the final result:P(0<X<2)+P(X=3)=0.07+0.01=0.08P(0 < X < 2) + P(X = 3) = 0.07 + 0.01 = 0.08P(0<X<2)+P(X=3)=0.07+0.01=0.08

This solution highlights the importance of carefully interpreting inequality expressions in the context of discrete random variables. Only integer values that satisfy the given conditions contribute to the probability sum.

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