Question

A random variable X has the following probability distribution 

X= x	1	2	3	4	5	6	7	8
P(X = x)	0.15	0.23	k	0.10	0.20	0.08	0.07	0.05

For the events E = {x/x is a prime number} and F = {x/x <4} then P(E ∪ F)

• A. 0.57
• B. 0.87
• C. 0.77
• D. 0.35

Answer 1

The Correct Option is C

To solve the problem, we need to find the probability of the union of events $E$ and $F$, where $E$ is the set of prime numbers and $F$ is the set of numbers less than 4, given a probability distribution for a random variable $X$.

1. Finding the Missing Probability:
The sum of probabilities for all possible values of $X$ must equal 1. 
The given probabilities are 0.15, 0.23, $k$, 0.10, 0.20, 0.08, 0.07, and 0.05. Thus:
$ 0.15 + 0.23 + k + 0.10 + 0.20 + 0.08 + 0.07 + 0.05 = 1 $
Summing the known probabilities:
$ 0.15 + 0.23 + 0.10 + 0.20 + 0.08 + 0.07 + 0.05 = 0.88 $
So:
$ 0.88 + k = 1 $
Solving for $k$:
$ k = 1 - 0.88 = 0.12 $

2. Defining the Events:
Event $E$ is the set of prime numbers:
$ E = \{ x | x \text{ is a prime number} \} = \{2, 3, 5, 7\} $
Event $F$ is the set of numbers less than 4:
$ F = \{ x | x < 4 \} = \{1, 2, 3\} $

3. Finding the Union:
The union of events $E$ and $F$ includes all elements in either $E$ or $F$:
$ E \cup F = \{1, 2, 3, 5, 7\} $

4. Calculating $P(E \cup F)$:
The probability of the union is the sum of the probabilities of the elements in $E \cup F$:
$ P(E \cup F) = P(1) + P(2) + P(3) + P(5) + P(7) $
Using the given probabilities ($P(1) = 0.15$, $P(2) = 0.23$, $P(3) = k = 0.12$, $P(5) = 0.20$, $P(7) = 0.07$):
$ P(E \cup F) = 0.15 + 0.23 + 0.12 + 0.20 + 0.07 $

Summing these:

$ 0.15 + 0.23 = 0.38 $
$ 0.38 + 0.12 = 0.50 $
$ 0.50 + 0.20 = 0.70 $
$ 0.70 + 0.07 = 0.77 $
Thus:
$ P(E \cup F) = 0.77 $

Final Answer:
The probability of the union of events $E$ and $F$ is $0.77$.