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Zorluk: OrtaProbability of Independent, Dependent, and Mutually Exclusive Events

Two software security tools, Tool X and Tool Y, operate independently to scan code repositories for vulnerabilities. The probability that Tool X detects a specific type of security flaw is 0.800.80, and the probability that Tool Y detects the same flaw is 0.750.75. What is the probability that exactly one of the two tools detects the flaw?

Cevap: 0.35

Cevap

The probability that exactly one of the two tools detects the flaw is 0.350.35.
The scenario requires finding the probability that exactly one tool detects the flaw. For independent events XX and YY, 'exactly one' consists of two mutually exclusive events: (1) Tool X succeeds while Tool Y fails, which has probability 0.80×(1−0.75)=0.80×0.25=0.200.80 \times (1 - 0.75) = 0.80 \times 0.25 = 0.20, and (2) Tool Y succeeds while Tool X fails, which has probability 0.75×(1−0.80)=0.75×0.20=0.150.75 \times (1 - 0.80) = 0.75 \times 0.20 = 0.15. Summing these mutually exclusive probabilities gives 0.20+0.15=0.350.20 + 0.15 = 0.35. Alternatively, one can subtract the probability of both tools succeeding (0.80×0.75=0.600.80 \times 0.75 = 0.60) from the probability of at least one tool succeeding (0.80+0.75−0.60=0.950.80 + 0.75 - 0.60 = 0.95), yielding 0.95−0.60=0.350.95 - 0.60 = 0.35.

Adım Adım Çözüm

1
Determine the complement probabilities for each independent tool failing to detect the flaw.
P(Not X)=1−0.80=0.20P(\text{Not X}) = 1 - 0.80 = 0.20 and P(Not Y)=1−0.75=0.25P(\text{Not Y}) = 1 - 0.75 = 0.25.
The probability of an event not occurring is equal to 1 minus the probability that it occurs.
2
Calculate the joint probability of Tool X detecting the flaw and Tool Y failing to detect it.
P(X and Not Y)=0.80×0.25=0.20P(\text{X and Not Y}) = 0.80 \times 0.25 = 0.20.
Because the tools operate independently, the joint probability is the product of their individual probabilities.
3
Calculate the joint probability of Tool Y detecting the flaw and Tool X failing to detect it.
P(Y and Not X)=0.75×0.20=0.15P(\text{Y and Not X}) = 0.75 \times 0.20 = 0.15.
Tool independence allows multiplying the individual probabilities of detection and non-detection.
4
Sum the probabilities of the two mutually exclusive scenarios representing 'exactly one tool detects the flaw'.
P(Exactly One)=0.20+0.15=0.35P(\text{Exactly One}) = 0.20 + 0.15 = 0.35.
The events 'X only' and 'Y only' cannot happen simultaneously, so their probabilities add directly.

Anahtar Kavram

Independence and Mutual Exclusivity Rules in Compound Probability
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Konular

Probability of Independent, Dependent, and Mutually Exclusive Events

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