A random variable X has the probability distr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The prime numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$ are $2, 3, 5, 7$.$P(E) = P(X=2) + P(X=3) + P(X=5) + P(X=7) = 0.23 + 0.12 + 0.20 + 0.07 =

Vedclass · Accepted Answer

$0.77$

Vedclass · Answer

(B) The prime numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$ are $2, 3, 5, 7$.$P(E) = P(X=2) + P(X=3) + P(X=5) + P(X=7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62$.The event $F = \{X $P(F) = P(X=1) + P(X=2) + P(X=3) = 0.15 + 0.23 + 0.12 = 0.50$.The intersection $E \cap F$ consists of values that are both prime and less than $4$,which are $\{2, 3\}$.$P(E \cap F) = P(X=2) + P(X=3) = 0.23 + 0.12 = 0.35$.Using the addition theorem of probability,$P(E \cup F) = P(E) + P(F) - P(E \cap F)$.$P(E \cup F) = 0.62 + 0.50 - 0.35 = 0.77$.

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

$A$ random variable $X$ has the probability distribution:
$X$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
$P(X)$ $0.15$ $0.23$ $0.12$ $0.10$ $0.20$ $0.08$ $0.07$ $0.05$

For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$,then $P(E \cup F)$ is

Similar Questions

If $P(A \cup B) = \frac{2}{3}$,$P(A \cap B) = \frac{1}{6}$,and $P(A) = \frac{1}{3}$,then which of the following is true?

$A, B, C$ are three events,one of which must and only one can happen. The odds in favour of $A$ are $4:6$,and the odds against $B$ are $7:3$. Then,the odds against $C$ are:

If $P(A) = 2/3$,$P(B) = 1/2$ and $P(A \cup B) = 5/6$,then events $A$ and $B$ are

If the odds against solving a question by three students are $2:1$,$5:2$,and $5:3$ respectively,then the probability that the question is solved by only one student is

For two events $A$ and $B$,if $P(A \cup B) = 5/6$,$P(A^c) = 5/6$,and $P(B) = 2/3$,then $A$ and $B$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$ random variable $X$ has the probability distribution:$X$$1$$2$$3$$4$$5$$6$$7$$8$$P(X)$$0.15$$0.23$$0.12$$0.10$$0.20$$0.08$$0.07$$0.05$For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$,then $P(E \cup F)$ is