Probability of transmitting a signal through a network of transmitters.
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We have a network of four transmitters $A$, $B$, $C$ and $D$. What is the probability of transmitting a signal through the network if all transmitters work independently and the probabilities of transmitting a signal by each transmitter are $0.7$, $0.8$, $0.9$ adn $0.6$, respectively.
My take is the following. The probability of transmitting a signal through $C$ and then $D$ is $P(C cap D) = 0.9 cdot 0.6 = 0.54$. The event "a signal transmitted through $A$ or $B$" consists of three disjoint events: $overline A cap B$, $A cap overline B$ and $A cap B$. We have $P(A cup B) = P(overline A cap B) + P(A cap overline B) + P(A cap B) = 0.3 cdot 0.8 + 0.7 cdot 0.2 + 0.7 cdot 0.8 = 0.94$.
So the probability of transmitting a signal through the entire network is:
$$P(A cup B) cdot P(C cap D) = 0.94 cdot 0.54 = 0.5076$$. Is that correct?
probability independence
asked Dec 20 '18 at 23:59
dkalocinskidkalocinski
526
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$begingroup$
We have a network of four transmitters $A$, $B$, $C$ and $D$. What is the probability of transmitting a signal through the network if all transmitters work independently and the probabilities of transmitting a signal by each transmitter are $0.7$, $0.8$, $0.9$ adn $0.6$, respectively.
My take is the following. The probability of transmitting a signal through $C$ and then $D$ is $P(C cap D) = 0.9 cdot 0.6 = 0.54$. The event "a signal transmitted through $A$ or $B$" consists of three disjoint events: $overline A cap B$, $A cap overline B$ and $A cap B$. We have $P(A cup B) = P(overline A cap B) + P(A cap overline B) + P(A cap B) = 0.3 cdot 0.8 + 0.7 cdot 0.2 + 0.7 cdot 0.8 = 0.94$.
So the probability of transmitting a signal through the entire network is:
$$P(A cup B) cdot P(C cap D) = 0.94 cdot 0.54 = 0.5076$$. Is that correct?
probability independence
asked Dec 20 '18 at 23:59
dkalocinskidkalocinski
526
$endgroup$
add a comment |
$begingroup$
We have a network of four transmitters $A$, $B$, $C$ and $D$. What is the probability of transmitting a signal through the network if all transmitters work independently and the probabilities of transmitting a signal by each transmitter are $0.7$, $0.8$, $0.9$ adn $0.6$, respectively.
My take is the following. The probability of transmitting a signal through $C$ and then $D$ is $P(C cap D) = 0.9 cdot 0.6 = 0.54$. The event "a signal transmitted through $A$ or $B$" consists of three disjoint events: $overline A cap B$, $A cap overline B$ and $A cap B$. We have $P(A cup B) = P(overline A cap B) + P(A cap overline B) + P(A cap B) = 0.3 cdot 0.8 + 0.7 cdot 0.2 + 0.7 cdot 0.8 = 0.94$.
So the probability of transmitting a signal through the entire network is:
$$P(A cup B) cdot P(C cap D) = 0.94 cdot 0.54 = 0.5076$$. Is that correct?
probability independence
asked Dec 20 '18 at 23:59
dkalocinskidkalocinski
526
$endgroup$
We have a network of four transmitters $A$, $B$, $C$ and $D$. What is the probability of transmitting a signal through the network if all transmitters work independently and the probabilities of transmitting a signal by each transmitter are $0.7$, $0.8$, $0.9$ adn $0.6$, respectively.
My take is the following. The probability of transmitting a signal through $C$ and then $D$ is $P(C cap D) = 0.9 cdot 0.6 = 0.54$. The event "a signal transmitted through $A$ or $B$" consists of three disjoint events: $overline A cap B$, $A cap overline B$ and $A cap B$. We have $P(A cup B) = P(overline A cap B) + P(A cap overline B) + P(A cap B) = 0.3 cdot 0.8 + 0.7 cdot 0.2 + 0.7 cdot 0.8 = 0.94$.
So the probability of transmitting a signal through the entire network is:
$$P(A cup B) cdot P(C cap D) = 0.94 cdot 0.54 = 0.5076$$. Is that correct?
probability independence
probability independence
asked Dec 20 '18 at 23:59
dkalocinskidkalocinski
526
asked Dec 20 '18 at 23:59
dkalocinskidkalocinski
526
asked Dec 20 '18 at 23:59
dkalocinskidkalocinski
526
asked Dec 20 '18 at 23:59
dkalocinskidkalocinski
526
asked Dec 20 '18 at 23:59
dkalocinskidkalocinski
526
526
add a comment |
add a comment |
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Yes that is correct
Another calculation could be $$Pleft(overline{(overline A cap overline B)} cap C cap Dright)=(1-(1-0.7)times (1-0.8))times 0.9times0.6 = 0.5076$$
answered Dec 21 '18 at 0:11
HenryHenry
101k482169
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$begingroup$
Yes that is correct
Another calculation could be $$Pleft(overline{(overline A cap overline B)} cap C cap Dright)=(1-(1-0.7)times (1-0.8))times 0.9times0.6 = 0.5076$$
answered Dec 21 '18 at 0:11
HenryHenry
101k482169
$endgroup$
add a comment |
$begingroup$
Yes that is correct
Another calculation could be $$Pleft(overline{(overline A cap overline B)} cap C cap Dright)=(1-(1-0.7)times (1-0.8))times 0.9times0.6 = 0.5076$$
answered Dec 21 '18 at 0:11
HenryHenry
101k482169
$endgroup$
add a comment |
$begingroup$
Yes that is correct
Another calculation could be $$Pleft(overline{(overline A cap overline B)} cap C cap Dright)=(1-(1-0.7)times (1-0.8))times 0.9times0.6 = 0.5076$$
answered Dec 21 '18 at 0:11
HenryHenry
101k482169
$endgroup$
Yes that is correct
Another calculation could be $$Pleft(overline{(overline A cap overline B)} cap C cap Dright)=(1-(1-0.7)times (1-0.8))times 0.9times0.6 = 0.5076$$
answered Dec 21 '18 at 0:11
HenryHenry
101k482169
answered Dec 21 '18 at 0:11
HenryHenry
101k482169
answered Dec 21 '18 at 0:11
HenryHenry
101k482169
answered Dec 21 '18 at 0:11
HenryHenry
101k482169
101k482169
add a comment |
add a comment |
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