Question

22. Random numbers Let $Y$be a number between$0$ and $1$ produced by a random number generator. Assuming that the random variable $Y$ has a uniform distribution, find the following probabilities:
(a) $P(Y\le 0.4)$
(b) role="math" localid="1649593908596" $P(Y<0.4)$
(c) role="math" localid="1649593922659" $P(0.1<Y\le 0.15or0.77\le Y<0.88$

Short answer

(a) The probability for $P(Y\le 0.4)$is $0.40$

(b) The probability for $P(Y<0.4$ is $0.40$

(c) The probability for role="math" localid="1649594539701" $P(0.1<Y\le 0.15or0.77\le Y<0.88)$is $0.16$.

Step-by-step solution

Part (a) Step 1: Given information 

To find the probability for $P(Y\le 0.4)$.

Part (a) Step 2: Explanation 

The variable$\mathrm{Y}$ is following the uniform distribution.
The probability for$P(Y\le 0.4)$can be determined as:
$$\begin{gathered} P(Y\le 0.4)=(0.4-0)\times 1 \\ =0.40 \end{gathered}$$

Part (b) Step 1: Given information

To find the probability for $P(Y<0.4)$

Part (c) Step 2: Explanation 

The variable$\mathrm{Y}$ is following the uniform distribution.
The probability for$P(Y<0.4)$ can be determined as:
$$\begin{gathered} P(Y<0.4)=(0.4-0)\times 1 \\ =0.40 \end{gathered}$$

Part (c) Step 1: Given information 

To find the probability for $P(0.1<Y\le 0.15or0.77\le Y<0.88)$.

Part (c) Step 2: Explanation 

The probability for$P(0.1<Y\le 0.15or0.77\le Y<0.88)$ can be determined as:
$$\begin{gathered} P(0.1<Y\le 0.15)=(0.15-0.1)\times 1 \\ =0.05 \end{gathered}$$
$$\begin{gathered} P(0.77\le Y\le 0.88)=(0.88-0.77)\times 1 \\ =0.11 \end{gathered}$$

Then,

$$\begin{gathered} P(0.1<Y\le 0.15or0.77\le Y\le 0.88)=0.05+0.11 \\ =0.16 \end{gathered}$$