In fact, the probability of obtaining three successes and two failures in five trials is p3(1-p)2 for each of the different ways this could occur, i.e. SSSFF, SSFSF, ... etc.
Consider n Bernoulli trials.
Assume the probability of success (S) is the same for all trials, P(S) = p.
Assume trials are independent so the probability for any given combination of successes and failures, e.g. for 5 trials, the probability of the outcome SSFSF, can be obtained by multiplying the probabilities for each trial outcome.
e.g. P(SSFSF) = p.p.(1-p).p.(1-p) = p3(1-p)2
In fact, the probability of obtaining three successes and two failures
in five trials is p3(1-p)2 for each of the different
ways this could occur, i.e. SSSFF, SSFSF, ... etc.
The number of distinct "arrangements" of 3 successes and 2
failures can be easily calculated using the binomial coefficient 
where n is the number of
trials and x is the number of successes required.
The binomial coefficient
(read as "n choose x") is defined as
Let X be the RV equal to the total number of successes in n trials.
To calculate the probability of obtaining x successes, it can be shown that
where the minimum number of successes is 0 and the maximum is n.
The distribution of the count of successes is called the binomial distribution with two parameters, n and p, required to determine P(X=x). We say X ~ B(n,p)
Then X has binomial distribution with n=3 and p=0.5, outcome win(W) or Lose(L) on each trial.
(This is abbreviated to X ~ B(3,0.5))
What is the probability that the team will win exactly 2 games?
P(X=2) = P(WWL) + P(WLW) + P(LWW)
= 3/8 from first principles (draw a tree diagram)
or using the formula for binomial probabilities,
Binomial distributions are used to model situations which can be thought of as repeated independent "trials" each with only 2 possible outcomes. We will use them later to make statistical inferences about proportions.
A quality control inspection system requires that from each batch of items a sample of 10 is selected and tested. If 2 or more of the sample are defective the whole batch is rejected.
If the probability of an item being defective is 0.05
Let X be RV = number of defectives in the sample of n = 10 items.
Assume X ~ Binomial (10, 0.05)
| ... Previous page | Next page ... |