Two independent security systems, X and Y, operate in a facility. The probability that sys… — JAMB UTME

Two independent security systems, $X$ and $Y$ , operate in a facility. The probability that system $Y$ detects an intrusion is $0.20$ greater than the probability that system $X$ detects it. If the probability that at least one of the systems detects an intrusion is $0.92$ , what is the probability that exactly one of the systems detects the intrusion?

$0.44$ Cevap
B
$0.48$
C
$0.52$
D
$0.32$

Cevap

The probability that exactly one of the systems detects the intrusion is

0.44

The correct answer is

0.44

. Setting

P(X) = p

and

P(Y) = p + 0.20

, we use the independence rule

P(X \cap Y) = P(X)P(Y)

in the addition law

P(X \cup Y) = P(X) + P(Y) - P(X)P(Y) = 0.92

. Solving

p^2 - 1.80p + 0.72 = 0

gives

P(X) = 0.60

and

P(Y) = 0.80

. The probability of both detecting the intrusion is

0.60 \times 0.80 = 0.48

. Subtracting the probability of both from the probability of at least one (

0.92 - 0.48

) gives

0.44

for exactly one system detecting the intrusion.

Adım Adım Çözüm

Define variables for the individual probabilities

Let

P(X) = p

. Then

P(Y) = p + 0.20

System

Y

's probability is given as

0.20

greater than System

X

's probability.

Apply the addition law for independent events

P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = p + (p + 0.20) - p(p + 0.20) = 0.92

Since

X

and

Y

are independent,

P(X \cap Y) = P(X)P(Y)

Solve the quadratic equation for

p

2p + 0.20 - p^2 - 0.20p = 0.92 \implies -p^2 + 1.80p - 0.72 = 0 \implies p^2 - 1.80p + 0.72 = 0

. Factoring yields

(p - 0.60)(p - 1.20) = 0

. Since

p \le 1

p = 0.60

Probability values cannot exceed

1

, so

p = 0.60

is the valid root.

Calculate individual probabilities and the probability of both occurring

P(X) = 0.60

P(Y) = 0.80

, and

P(X \cap Y) = 0.60 \times 0.80 = 0.48

These are needed to evaluate the compound probability of exactly one event occurring.

Calculate the probability that exactly one system detects the intrusion

P(\text{exactly one}) = P(X \cup Y) - P(X \cap Y) = 0.92 - 0.48 = 0.44

The probability of exactly one event occurring is the probability of at least one minus the probability of both.

Anahtar Kavram

Probability laws for independent and compound events

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