Bernoulli’s trials

Bernoulli’s trials refer to repeated random experiments when there is a possibility of either success or failure. Bernoulli’s trials follow the binomial expansion (q + p) n whereq is the probability of failure, p is the probability of success, and q + p = 1.

Frequently the probability generating function for these n trials is written in the form (q + p)n = nC0 qn + nC1 qn -1p + nC2 qn – 2p2 + nC3 qn – 3p3 + …

which will give:

Probability of no successes + probability of one success + probability of two successes + probability of three successes, etc.

The general terms of this expansion can also be written as nCk qn - kpk.

The variable X is called the random variable of the binomial distribution. It is often referred to as a binomial variable in the formula: P(X = k) = nCk qn -krpk.

For example, P(X = O) represents the probability of no successes.

The expansion of(q + p)ncould, therefore, be expressed in the form

P(X = 0) + P(X = 1) + P(X = 2) + … + P(X = n).

Negative probability

Outcomes from Bernoulli’s trials have also been refereed to as negative binomial probabilities because they represent a special case in which the probability of failure q is written first in the generation function (q + p)n. Terms are expanded until the probability of k'th success occurs using the formula for the general term nCk qn - kpk.

Worked out problem

Question 1

Find the probability of getting exactly five successes.

If X is the probability of successes in the trials then the probability of exactly five successes could be written asP(X = 5) and the probability generating function is as following:

Question 2

Find the probability of getting at least five successes.

For at least five successes X can have the values 5, 6, 7, or 8.

ie X ≥ 5 where X belongs to the set {0, 1, 2, 3, … 8}

Therefore, the probability of at least five successes could be written as:

P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

Question 3

Find the probability of getting as many successes as there are failures.

For “as many successes as there are failures” we require

X = 4

For similar questions check:
  • 1995 HSC Extension 1Exam Paper – Question 5b.
  • 1999 HSC Extension 1Exam Paper – Question 2b.
  • 2002 HSC Extension 1 Exam Paper – Question 4a.
  • 2003 HSC Extension 1 Exam Paper – Question 3c.
  • 2004 HSC Extension1 Exam Paper – Question 4c.
  • 2005 HSC Extension 1 Exam Paper – Question 6a.
Link:

Activity

/ Life is so slow in Slowville that traffic lights are either red or green. The probability of passing (P) through a green light is always 0.5 and the probability of being stopped (S) at a red light is also 0.5 .

If you need to go through three traffic lights in Slowville, find the probability of passing through:

a) two green lights

b) three green lights.

Find also the probability of being stopped:

c) at least twice

d) at three traffic lights.

Hint: (P + S)3 where P is the probability of passing through and S is the probability of being stopped.

Feedback

The generating binomial probability function is:

(P + S)3 = P3 + 3P2S + 3PS2 + S3

a) P(two green lights) = 3P2S

= 3(0.5)2(0.5)

= 0.375

b) P(three green lights) = P3

= (0.5)3

= 0.125

c) P(at least twice) = 3PS2 + S3

= 3(0.5)(0.5)2 + (0.5)3

= 0.375 + 0.125

= 0.5

d) P(three traffic lights) = S3

= (0.5)3

= 0.125

1

2004