Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro
2 answers:
The question appears to be incomplete. Here’s the complete inquiry:
Samir is quite skilled with the gun. When he targets a specific aim at the shooting range, he has a 0.95 probability of striking it. On one occasion, Samir sets out to shoot 10 targets consecutively.
If he has the same chance of hitting each of the 10 targets, what is the likelihood that he will miss at least one?
Response:
40.13%
Step-by-step breakdown:
Let 'A' represent the event of successfully hitting all targets in 10 trials.
The complement of 'A' is
Now, since Samir has a consistent probability of hitting each target at 0.95.
Now,
We know that the combined probability of an event and its complement equals 1.
<pThus,Consequently, the probability that he misses at least one target among 10 attempts is 40.13%.
You might be interested in
Try this approach (refer to the attachment, which outlines three steps).
Answer:
The anticipated number of tests required to identify 680 acceptable circuits is 907.
Step-by-step explanation:
For any circuit, there are two potential results: it either passes the test or it fails. The likelihood of passing is independent between circuits. Therefore, we apply the binomial probability distribution to address this scenario.
Binomial probability distribution
This distribution calculates the chance of obtaining exactly x successes across n trials, where x has only two possible outcomes.
To find the expected number of trials to achieve r successes with a probability p, the formula is given by:
Circuits from a specific factory pass a certain quality evaluation with a probability of 0.75.
Thus, to determine the expected number of tests needed for 680 acceptable circuits, let’s denote this as E where r = 680.
The expected number of tests necessary to find 680 acceptable circuits is 907.