There are 4 runners on the New High School team. The team is planning to participate in a race in which each runner runs a mile. The team time is the sum of the individual times for the 4 runners. Assume that the individual times of the 4 runners are all independent of each other. The individual times, in minutes, of the runners in similar races are approximately normally distributed with the following means and standard deviations.
2 outcomes: each observation can be categorized as a “success” or “failure”
N: Fixed # of observations
I: Observations are independent
P: The probability is the same for each observation
Coin toss to see which of the 2 football teams gets the choice of kicking off or receiving to begin the game
A couple prepares for their first child
Although sometimes a problem is not strictly independent and therefore not a binomial setting, the situation is close enough for practical purposes.
An engineer selects an SRS of 10 switches from a large shipment for detailed inspection. Unknown to the engineer, 10% of them fail to meet the specifications. What is the probability that no more than 1 of the 10 switches in the sample fail inspection?
(Note: Assuming that a defective switch is drawn first (p=0.1), the probability for the second switch being defective changes to p=0.0999. For practical purposes, this behaves like a binomial setting even if the condition of independence does not strictly hold).
P(X=0): Binompdf(10,.1,0) = .3486784401
P(X=1): Binompdf(10, .1, 1) = .387420489
= .3487 + .3874 = .7361
P(X 1) = binomcdf(10, .1, 1) = .736098903 (same answer, less work)
Same answer, less work:
P(X 7) = binomcdf(12, .75, 7) = .1576436761
Example: Each child born to a set of parents has probability .25 of having blood type O. If these parents have 5 kids, what is the probability that exactly 2 of them have type O blood?
1st way: List all 10 possible outcomes: SSFFF, SFSFF, SFFSF, SFFFS, FSSFF, FSFSF, FSFFS, FFSSF, FFSFS, FFFSS, each with the same probability (for ex, the 1st: SSFFF = (.25)(.25)(.75)(.75)(.75))
2nd way: Using the binomial formula!