Respuesta :
Step-by-step explanation:
Probability, P = [tex]\frac{no. of favourable outcomes}{Total no. of possible outcomes}[/tex]
Out of 8846 heart pacemaker malfunctions cases, caused by firmware cases are 375
then, no. of cases not caused by firmware are:
8846 - 375 = 8471
Probability for three different pacemakers respectively is given by:
P(1) = [tex]\frac{8471}{8846}[/tex]
we select the next malfunctioned pacemaker from the remaining i.e., out of 8471, excluding the chosen malfunctioned pacemaker
P(2|1) = [tex]\frac{8470}{8845}[/tex]
Therefore, the events are not independent of each other
Now, if the selection is without replacement, then
P(3|1 & 2) = [tex]\frac{8469}{8844}[/tex]
Now, by general multiplication rule(as the events are not independent):
P(none of the 3 are caused by malfunction) =
[tex]\frac{8471}{8846}\times\frac{8470}{8845}\times\frac{8469}{8844}[/tex]
= 0.9019
P(none of the 3 are caused by malfunction) = 90.19%
The probability is high, therefore the whole batch will be accepted.
Probability of an event represents the chances of that event to occur.
- The probability of the entire batch to get selected is 0.879
- Yes, this procedure is likely to result in the entire batch being accepted.
What is the multiplication rule of probability for independent events?
Suppose there are n mutually independent events.
The probability of their simultaneously occurrence is given as
[tex]P(A_1 \cap A_2 \cap ... \cap A_n) = P(A_1) \times P(A_2) \times ... \times P(A_n)[/tex]
(This is true only if all those events are mutually independent).
The probability of getting a failure(malfunction) in a randomly selected pacemaker for the given case is
[tex]\dfrac{375}{8846} \approx 0.042[/tex]
The probability of a pacemaker working fine = 1- its failure probability = 1 - 0.042 = 0.958
Since each of the 3 selected pacemakers are independent for their failure or success of each other, thus, if we have:
[tex]A_1[/tex] = Event of properly working of first pacemaker
[tex]A_2[/tex] = Event of properly working of second pacemaker
[tex]A_3[/tex] = Event of properly working of third pacemaker
Then, as all three are independent for their working from each other, thus,
[tex]P(A_1 \cap A_2 \cap A_3) = P(A_1) \times P(A_2)\times P(A_3) = (0.958)^3 \approx 0.879\\\\\begin{aligned}{P(\text{batch getting selected}) &= P(\text{All three pacemaker working})\\& = P(A_1 \cap A_2 \cap A_3) \\&\approx 0.879\\\end{aligned}[/tex]
Thus, as this probability is high, thus, this procedure is likely to result in the entire batch being accepted.
Thus,
- The probability of the entire batch to get selected is 0.879
- Yes, this procedure is likely to result in the entire batch being accepted.
Learn more about multiplication rule of probability here:
https://brainly.com/question/14399918
