Question

21. Random numbers Let $X$be a number between $0$and $1$produced by a random number generator. Assuming that the random variable X has a uniform distribution, find the following probabilities:
(a) $P(X>0.49)$
(b) $P(X\ge 0.49)$
(c) $P(0.19\le X<0.37or0.84<X\le 1.27)$

Short answer

(a) The probability for $P(X>0.49)$is $0.51$.

(b) The probability for $P(X\ge 0.49)$is $0.51$.

(c) The probability for $P(0.19\le X<0.37or0.84<X\le 1.27)$is $0.34$.

Step-by-step solution

Part (a) Step 1: Given information 

Given in the question that, Consider $X$as a number between $0$and $1$created by a random number generator.

We have to And to find the probability for $P(X>0.49)$.

Part (a) Step 2: Explanation

The variable$X$ has the uniform distribution.
Therefore, the value of $P(X>0.49)$by:

$$\begin{gathered} P(X>0.49)=(1-0.49)\times 1 \\ =0.51 \end{gathered}$$

Part (b) Step 1: Given information 

Let $X$be a number between $0$and $1$produced by a random number generator.

We have to find the probability for $P(X\ge 0.49)$.

Part (b) Step 2: Explanation 

The variable $X$has the uniform distribution.

Therefore, the value of $P(X\ge 0.49)$will be:

$$\begin{gathered} P(X\ge 0.49)=(1-0.49)\times 1 \\ =0.51 \end{gathered}$$

Part (c) Step 1: Given information 

Let $X$be a number between $0$and $1$ produced by a random number generator.

We have to find the probability for $P(0.19\le X<0.37or0.84<X\le 1.27)$.

Part (c) Step 2: Explanation 

Since, the probability for$P(0.19\le X<0.37or0.84<X\le 1.27)$can determine as:
$$\begin{gathered} P(0.19\le X<0.37)=(0.37-0.19)\times 1 \\ =0.18 \end{gathered}$$
$$\begin{gathered} P(0.84<X\le 1.27)=(1-0.84)\times 1 \\ =0.16 \end{gathered}$$

Then,

$$\begin{gathered} P(0.19\le X<0.37)=0.18+0.16 \\ =0.34 \\ or \\ P(0.84<X\le 1.27)=0.18+0.16 \\ =0.34 \end{gathered}$$