Repeated Trials in probability
Repeated trials refer to the outcomes of events that repeat one or more times. The probability of success (p) and failure (q) are used to determine the total probability of n successes in a trial.
For example,
The probability of tossing a coin thrice
We know that a coin has two possible outcomes, Head (H) and Tail (T). The probability of the occurrence of Head is P(H) =1/2. The probability of the occurrence of Tail is P(T) = 1/2. The probability of Head (H) and Tail (T) outcomes is equal because both have equal chances to occur.
The outcomes of tossing a coin thrice are (H, H, H), (H, H, T), (H, T, H), (T, H, H), (T, T, H), (T, H, T), (H, T, T), and (T, T, T).
The outcomes of getting two heads and one tail = (H, H, T), (H, T, H), (T, H, H)
Probability of each outcome will be (1/2 x 1/2 x 1/2) = (1/2)3
Total probability of getting two heads and one tail = 3 x (1/2)3
The outcomes of tossing a coin twice are (H, H), (H, T), (T, H), (T, T). Similarly, we can find the probability of various outcomes.
If a trial is repeated n time, the probability of ‘r’ successes and failures ‘n – r’ will be:
pr qn – r
Where,
p = probability of success
q = probability of failure
An experiment that repeats identically is known as repeated trials experiment in probability. A repeated trial in probability reduces the errors, increases the reliability, and accuracy of an experiment.
Binomial Distribution
It is concerned with the repeated trials where the success, failure, occurrence, or non-occurrence concepts are used. We know that the probability of ‘r’ successes in series of n trials is given by:
nCr pr qn – r
Where,
p = probability of success
q = probability of failure
n is the total number of trials
r refers to the success in series of n trials. The value of r lies between 0 and n.
The probability of 0 success is represented as: nC0 p0 qn – 0 = qn
The probability of 1 success is represented as: nC1 p1 qn – 1
The probability of 2 successes is represented as: nC2 p2 qn – 2
The probability of n success is represented as: nCn pn qn – n = pn
(nC0 = 1, nCn = 1, p0 = 1, q0 = 1)
The sum of probabilities is always 1
nC0 p0 qn – 0 + nC1 p1 qn – 1 + nC2 p2 qn – 2 + nCn pn qn – n
= qn + nC1 p1 qn – 1 + nC2 p2 qn – 2 + pn
= (qn + pn) = 1
Examples
Let’s discuss some numerical examples based on the repeated trials.
Example 1: The probability that a pencil manufactured by a company will be defective is 2/10. If 10 such pens are manufactured, find the probability (a) exactly three will be defective (b) atleast two will be defective (c) none is defective.
Solution:
Probability that a pencil is defective = 2/10 = 0.2
Probability that a pencil is not defective = (1 – 0.2) = 0.8
Here,
p = 0.2
q = 0.8
n = 10
a) Probability that exactly three will be defective is:
= nC3 p3qn – 3
= 10C3 (0.2)3 (0.8)7
b) Probability that atleast two will be defective is:
= 1 – [(probability that none is defective) + (probability that one is defective)]
= 1 – (10C0 (0.2)0 (0.8)10 + 10C1 (0.2)1 (0.8)9)
= 1 – ( (0.8)10 + 10C1 (0.2)1 (0.8)9)
c) Probability that none will be defective is:
= nC0 p0qn – 0
= 10C0 (0.2)0 (0.8)10
= (0.8)10
Example 2: Find the probability of getting 5 heads, when we toss a coin 15 times?
Solution: In a coin tossing experiment, there are two outcomes, Head and Tail. The probability that both can occur is equal to 1/2 or 0.5.
Given: The number of repeated trials = 15
The number of success trials of getting heads = 5
Probability of success = ½ = 0.5
Probability of failure = (1 – 0.5) = 0.5
Using the binomial distribution, the probability of ‘r’ successes in series of n trials is given by:
nCr pr qn – r
where,
n = 15
r = 6
p = 0.5
q = 0.5
Substituting the values, we get:
15C6 (0.5)6 (0.5)9
After solving, we get:
= 0.0139 (approx)
Example 3: What can be the cases one can expect 10 heads and 6 tails in 256 sets of 16 tosses of a coin?
Solution: A single toss of coin produces two outcomes, Head and Tail, which has equal chances to occur.
P (Head) = 1/2
P (Tail) = 1/2
By Binomial distribution, the probability of 10 heads and 6 tails in total 16 tosses of a coin will be:
P (X = 10) = 16C10 (1/2)10 (1/2)6
P (X = 10) = 16C10 (1/2)16
P (X = 10) = 16C10 (0.5)16
P (X = 10) = 16C10 (0.5)16
P (X = 10) = 0.122
The expected number of cases in 256 sets is:
256 x P (X = 10)
= 256 x 0.122
= 31.28
Thus, the cases one can expect 10 heads and 6 tails in 256 sets of 16 tosses of a coin are 31.28.
Example 4: What can be the cases one can expect 4 heads and 2 tails in 16 sets of 6 tosses of a coin?
Solution: A single toss of coin produces two outcomes, Head and Tail, which has equal chances to occur.
P (Head) = 1/2
P (Tail) =1/2
By Binomial distribution, the probability of 4 heads and 2 tails in total 6 tosses of a coin will be:
P (X = 4) = 6C4 (1/2)4 (1/2)2
P (X = 4) = 6C4 (1/2)6
P = 0.234
The expected number of cases in 16 sets is:
16 x P (X = 4)
= 16 x 0.234
= 3.75
Thus, the cases one can expect 4 heads and 2 tails in 16 sets of 6 tosses of a coin are 3.75.
Example 5: In sampling several parts manufactured by a machine, the mean number of defective items in a sample of 20 is 2. Out of 2000 samples, how many are expected to contain atleast 2 defective parts?
Solution: The mean number of defectives = 2
Mean = np
Where,
n is the number of items
p is the probability that a sample is defective
np = 2
20p = 2
P = 2/20
= 0.1
The probability that a sample is defective = 0.1
The probability that a sample is not defective = (1 – 0.1) = 0.9
The probability of atleast two defective in a sample of 20 is:
1 – (Probability that none is defective + probability that one is defective)
= 1 – (20C0 (0.9)20 + 20C1 (0.1) (0.9)19)
= 1 – ((0.9)20 + 20 (0.1) (0.9)19)
= 1 – (0.3917)
= 0.608
Thus, number of samples having atleast two defective out of 2000 samples is:
2000 x 0.608
= 1216.5