Question

The probability of a successful optical alignment in the assembly of an optical data storage product is p = 0.6. Assume the trials are independent. Round your answers to four decimal places (e.g. 98.7654). (a) What is the probability that the 1st successful alignment requires exactly 4 trials? (b) What is the probability that the 1st successful alignment requires at most 4 trials? (c) What is the probability that the 1st successful alignment requires at least 4 trials?

Answer 1

Answer:

a) $P(X=4)=(1-0.6)^{4-1} 0.6 = 0.0384$

b) $P(X\leq 4)=0.6+0.24+0.096+0.0384=0.9744$

c) $P(X\geq 4)=1-P(X$

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

Part a

For this case we want this probability

$P(X=4)=(1-0.6)^{4-1} 0.6 = 0.0384$

Part b

For this case we want this probability:

$P(X\leq 4)=P(X=1)+P(X=2)+P(X=3)+P(X=4)$

If we find the individual probabilities we got:

$P(X=1)=(1-0.6)^{1-1} 0.6 = 0.6$

$P(X=2)=(1-0.6)^{2-1} 0.6 = 0.24$

$P(X=3)=(1-0.6)^{3-1} 0.6 = 0.096$

$P(X=4)=(1-0.6)^{4-1} 0.6 = 0.0384$

And replacing we have:

$P(X\leq 4)=0.6+0.24+0.096+0.0384=0.9744$

Part c

For this case at least 4 trials means that the random variable X needs to be 4 or more

$P(X\geq 4)=1-P(X$

And we found already the probabilities P(X=1),P(X=2) and P(X=3) so we just need to replace:

$P(X\geq 4)=1-P(X$