A random variable x has probability distribution:X=xi12345678P(X=xi0.150.230.120.100.200.080.070.05For the event E={ X is a prime number] and F={x<4}then P(E∪F)=

p(E)=0.23+0.12+0.20+0.07=0.62p(F)=p(x<4)=0.15+0.23+0.12=0.50p(E∩F)=p(x=2)+p(x=3)=0.23+0.12=0.35p(E∪F)=0.62+0.50−0.35=0.77

A random variable x has probability distribution:X=xi12345678P(X=xi0.150.230.120.100.200.080.070.05For the event E={ X is a prime number] and F={x<4}then P(E∪F)=

p(E)=0.23+0.12+0.20+0.07=0.62p(F)=p(x<4)=0.15+0.23+0.12=0.50p(E∩F)=p(x=2)+p(x=3)=0.23+0.12=0.35p(E∪F)=0.62+0.50−0.35=0.77

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A random variable x has probability distribution:
$X =$
$x_{i}$
1
2
3
4
5
6
7
8
$\begin{array}{l} P (X \\ = x_{i} \end{array}$
0.15
0.23
0.12
0.10
0.20
0.08
0.07
0.05
For the event $E =$ { X is a prime number] and $F = {x < 4}$ then $P (E \cup F) =$

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0.35

0.50

0.87

0.77

answer is B.

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Detailed Solution

$p (E) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62$

$p (F) = p (x < 4)$

$= 0.15 + 0.23 + 0.12 = 0.50$

$p (E \cap F) = p (x = 2) + p (x = 3)$

$= 0.23 + 0.12 = 0.35$

$p (E \cup F) = 0.62 + 0.50 - 0.35 = 0.77$

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A random variable x has probability distribution:X=xi12345678P(X=xi0.150.230.120.100.200.080.070.05For the event E={ X is a prime number] and F={x<4}then P(E∪F)=

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A random variable x has probability distribution:
$X =$
$x_{i}$
1
2
3
4
5
6
7
8
$\begin{array}{l} P (X \\ = x_{i} \end{array}$
0.15
0.23
0.12
0.10
0.20
0.08
0.07
0.05
For the event $E =$ { X is a prime number] and $F = {x < 4}$ then $P (E \cup F) =$