At peak periods, 15% of attempted log-ins to an email service fail. Log-ins attempts are independent and each has the same probability of failing. Ann logs in repeatedly until she succeeds.
A. find the probability that Ann needs to log in at least four times before she succeeds.
B. What is a standard deviation for number of log-ins needed to succeed
Thank you!!
A. The probability that Ann needs to log in at least four times before she succeeds is 0.5062.
B. The standard deviation for the number of log-ins needed to succeed is 1.5206.
To solve this problem, we need to use the concept of geometric probability, which is used to calculate the probability of a specified number of independent trials until success.
A. To find the probability that Ann needs to log in at least four times before she succeeds, we need to calculate the probability of failing on the first three attempts and succeeding on the fourth attempt.
The probability of failing on the first attempt is 0.15 (15% failure rate), and the probability of succeeding on the fourth attempt is (1 - 0.15) = 0.85 (85% success rate).
Since each log-in attempt is independent, we can multiply the probabilities together:
P(failure on the first try) * P(failure on the second try) * P(failure on the third try) * P(success on the fourth try) = 0.15 * 0.15 * 0.15 * 0.85
= 0.0030375
So, the probability that Ann needs to log in at least four times before she succeeds is approximately 0.0030375.
B. To calculate the standard deviation for the number of log-ins needed to succeed, we need to use the formula:
Standard Deviation = √(1 - p) / p
Where p is the probability of success (in this case, 0.85).
Standard Deviation = √(1 - 0.85) / 0.85
= √(0.15) / 0.85
= 0.387
So, the standard deviation for the number of log-ins needed to succeed is approximately 0.387.
To find the probability and standard deviation in this scenario, we can use the concept of geometric distribution.
A. Probability that Ann needs to log in at least four times before she succeeds:
The probability of success on each log-in attempt is 1 - 0.15 = 0.85 (since 15% fail).
Therefore, the probability of failure on each log-in attempt is 0.15.
We need to find the probability that Ann succeeds on her fourth log-in attempt or later. This is the same as finding the probability of at least 3 failures before the first success.
P(failure) = P(First failure) + P(Second failure) + P(Third failure)
P(failure) = 0.15 + (0.15 * 0.15) + (0.15 * 0.15 * 0.15)
P(failure) = 0.15 + 0.0225 + 0.003375
P(failure) ≈ 0.175875
Therefore, the probability that Ann needs to log in at least four times before she succeeds is approximately 0.175875 or 17.59%.
B. Standard deviation for the number of log-ins needed to succeed:
The formula for the standard deviation of a geometric distribution is:
σ = √[(1 - p) / p^2]
Where p is the probability of success on each attempt, which is 0.85, as calculated earlier.
σ = √[(1 - 0.85) / (0.85^2)]
= √[0.15 / 0.7225]
≈ √0.2076
≈ 0.4551
Therefore, the standard deviation for the number of log-ins needed to succeed is approximately 0.4551.