Respuesta :
A door alarm works in 72 out of 100 cases.
surveillance camera works in 68 out of 100 cases.
The probability that method A is used = 72/100
The probability that method B is used = 68/100
Since both method are used we find the probability of A and B
Probability ( A and B) = [tex] \frac{72}{100} [/tex] * [tex] \frac{68}{100} [/tex]
= [tex] \frac{306}{625} [/tex]
The probability of the effective screeningtechnique when both methods are used together is[tex]\boxed{{\mathbf{0}}{\mathbf{.4896}}}[/tex].
Further explanation:
It is given that the door alarm works in [tex]72[/tex] out of [tex]100[/tex] cases.
Probability is the chance of any event that is likely to occur. It is the ratio of number of favorable events and the total number of possible outcomes.
Since the door alarm works in [tex]72[/tex] out of [tex]100[/tex]cases, the probability of the door alarm that is likely to occur is calculated by taking the ratio of the favorable events [tex]72[/tex] and the total number of possible outcome [tex]100[/tex] as follows:
[tex]\begin{aligned}P\left( {{\text{door alarm}}} \right)&= \frac{{72}}{{100}} \\&=0.72\\ \end{aligned}[/tex]
Therefore, the probability of the door alarm is [tex]{\mathbf{0}}{\mathbf{.72}}[/tex].
It is given that the surveillance camera works in [tex]68[/tex] out of [tex]100[/tex] cases.
Since the surveillance camera works in [tex]68[/tex] out of [tex]100[/tex]cases, the probability of the surveillance camera that is likely to occur is calculated by taking the ratio of the favorable events [tex]68[/tex] and the total number of possible outcome [tex]100[/tex] as follows:
[tex]\begin{aligned} P\left( {{\text{surveillance camera}}} \right)&=\frac{{68}}{{100}} \\&=0.68\\\end{aligned}[/tex]
Therefore, the probability of the surveillance camera is [tex]{\mathbf{0}}{\mathbf{.68}}[/tex].
Now, when both methods that is door alarm or surveillance camera is used together, the probability of both events occurring together can be calculated by using a formula[tex]P\left( {A \cap B} \right) = P\left( A \right) \times P\left( B \right)[/tex].
Here, Event A is door alarm and Event B is surveillance camera.
The probability of the door alarm is [tex]{\mathbf{0}}{\mathbf{.72}}[/tex] that is[tex]P\left( A \right) = 0.72[/tex].
The probability of the surveillance camera is [tex]{\mathbf{0}}{\mathbf{.68}}[/tex]that is[tex]P\left( B \right) = 0.68[/tex].
Substitute [tex]0.72[/tex] for [tex]P\left( A \right)[/tex] and [tex]0.68[/tex] for [tex]P\left( B \right)[/tex]in equation [tex]P\left( {A \cap B} \right) = P\left( A \right) \times P\left( B \right)[/tex] to obtain theprobability of the effective screeningtechnique when both methods are used together.
[tex]\begin{aligned} P\left( {A \cap B} \right)&= 0.72 \times 0.68\\&=0.4896\\\end{aligned}[/tex]
Therefore, the probability is[tex]\boxed{{\mathbf{0}}{\mathbf{.4896}}}[/tex].
Thus, theprobability of the effective screeningtechnique when both methods are used together is[tex]\boxed{{\mathbf{0}}{\mathbf{.4896}}}[/tex].
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Answer Details:
Grade: Junior High School
Subject: Mathematics
Chapter: Probability
Keywords:expression, probability, screening technique, camera, , two methods,